Soal Persamaan Trigonometri Dasar dan Pembahasan

Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) akan berbagi kumpulan soal Persamaan Trigonometri Dasar berserta pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.

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Soal Persamaan Trigonometri Dasar No. 1

Himpunan penyelesaian dari persamaan: $\sin (3x-15^\circ )=\frac{1}{2}\sqrt{2}$ untuk $0^\circ \le x\le 180^\circ $ adalah …
A. $\{20^\circ ,140^\circ \}$
B. $\{50^\circ ,170^\circ \}$
C. $\{20^\circ ,50^\circ ,140^\circ \}$
D. $\{20^\circ ,50^\circ ,140^\circ ,170^\circ \}$
E. $\{20^\circ ,50^\circ ,140^\circ ,170^\circ ,200^\circ \}$

Penyelesaian: Lihat/Tutup

$\sin (3x-15^\circ )=\frac{1}{2}\sqrt{2}$
$\sin (3x-15^\circ )=\sin 45^\circ $
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=3x-15^\circ $ dan $g(x)=45^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}3x-15^\circ &= 45^\circ +k.360^\circ \\ 3x &= 45^\circ +15^\circ +k.360^\circ \\ 3x &= 60^\circ +k.360^\circ \\ x &= 20^\circ +k.120^\circ \end{align}$
$k=0\to x=20^\circ $
$k=1\to x=140^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}3x-15^\circ &= (180^\circ -45^\circ )+k.360^\circ \\ 3x-15^\circ &= 135^\circ +k.360^\circ \\ 3x &= 135^\circ +15^\circ +k.360^\circ \\ 3x &= 150^\circ +k.360^\circ \\ x &= 50^\circ +k.120^\circ \end{align}$
$k=0\to x=50^\circ $
$k=1\to x=170^\circ $
HP = $\{20^\circ ,50^\circ ,140^\circ ,170^\circ \}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 2

Nilai $x$ yang memenuhi persamaan $2\sin 2x+2\sin x=0$ dan $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,60^\circ ,90^\circ \}$
B. $\{60^\circ ,90^\circ ,120^\circ \}$
C. $\{90^\circ ,120^\circ ,150^\circ \}$
D. $\{120^\circ ,150^\circ ,240^\circ \}$
E. $\{120^\circ ,180^\circ ,240^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}2\sin 2x+2\sin x &= 0 \\ \sin 2x+\sin x &= 0 \\ \sin 2x &= -\sin x \\ \sin 2x &= \sin (-x) \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=2x$ dan $g(x)=-x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x &= -x+k.360^\circ \\ 3x &= k.360^\circ \\ x &= k.120^\circ \end{align}$
$k=0\to x=0^\circ $
$k=1\to x=120^\circ $
$k=2\to x=240^\circ $
$k=3\to x=360^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x &= 180^\circ +x+k.360^\circ \\ x &= 180^\circ +k.360^\circ \end{align}$
$k=0\to x=180^\circ $
HP = $\{0^\circ ,120^\circ ,180^\circ ,240^\circ ,360^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 3

Nilai $x$ yang memenuhi $\tan (3x+60^\circ )=\sqrt{3}$ adalah ….
A. $90^\circ $
B. $110^\circ $
C. $120^\circ $
D. $130^\circ $
E. $230^\circ $

Penyelesaian: Lihat/Tutup

$\tan (3x+60^\circ )=\sqrt{3}$
$\tan (3x+60^\circ )=\tan 60^\circ $
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=3x+60^\circ $ dan $g(x)=60^\circ $ maka:
$\begin{align}\color{blue}f(x) &\color{blue}= g(x)+k.180^\circ \\ 3x+60^\circ &= 60^\circ +k.180^\circ \\ 3x &= 60^\circ -60^\circ +k.180^\circ \\ 3x &= k.180^\circ \\ x &= k.60^\circ \end{align}$
$k=0\to x=0^\circ $
$k=1\to x=60^\circ $
$k=2\to x=120^\circ $
$k=3\to x=180^\circ $
$k=4\to x=240^\circ $
$k=5\to x=300^\circ $
$k=6\to x=360^\circ $
HP = $\{0^\circ ,60^\circ ,120^\circ ,180^\circ ,240^\circ ,300^\circ ,360^\circ \}$
Jawaban: C

Soal Persamaan Trigonometri Dasar No. 4

Himpunan penyelesaian $\sin 4x-\cos 2x=0$ untuk $0^\circ < x < 360^\circ $ adalah ….
A. $\{15^\circ ,45^\circ ,75^\circ ,135^\circ \}$
B. $\{135^\circ ,195^\circ ,225^\circ ,255^\circ \}$
C. $\{15^\circ ,45^\circ ,195^\circ ,225^\circ \}$
D. $\{15^\circ ,75^\circ ,195^\circ ,255^\circ \}$
E. $\{15^\circ ,45^\circ ,75^\circ ,135^\circ ,195^\circ ,225^\circ ,255^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

Perbandingan Trigonometri Sudut Berelasi, $\cos \alpha =\sin (90^\circ -\alpha )$.
$\begin{align}\sin 4x-\cos 2x &= 0 \\ \sin 4x &= \cos 2x \\ \sin 4x &= \sin (90^\circ -2x) \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=4x$ dan $g(x)=90^\circ -2x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}4x &= 90^\circ -2x+k.360^\circ \\ 4x+2x &= 90^\circ +k.360^\circ \\ 6x &= 90^\circ +k.360^\circ \\ x &= 15^\circ +k.60^\circ \\ \end{align}$
$k=0\to x=15^\circ $
$k=1\to x=75^\circ $
$k=2\to x=135^\circ $
$k=3\to x=195^\circ $
$k=4\to x=255^\circ $
$k=5\to x=315^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}4x &= (180^\circ -(90^\circ -2x))+k.360^\circ \\ 4x &= 90^\circ +2x+k.360^\circ \\ 4x-2x &= 90^\circ +k.360^\circ \\ 2x &= 90^\circ +k.360^\circ \\ x &= 45^\circ +k.180^\circ \end{align}$
$k=0\to x=45^\circ $
$k=1\to x=225^\circ $
HP = $\{15^\circ ,45^\circ ,75^\circ ,135^\circ ,195^\circ ,225^\circ ,255^\circ ,315^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 5

Perbandingan sudut A terkecil terhadap sudut A terbesar yang memenuhi persamaan $2\cos A+1=0$ untuk $0\le A\le 2\pi $ adalah ….
A. 1 : 1
B. 1 : 2
C. 2 : 3
D. 3 : 4
E. 4 : 5

Penyelesaian: Lihat/Tutup

$\begin{align}2\cos A+1 &= 0 \\ 2\cos A &= -1 \\ \cos A &= -\frac{1}{2} \end{align}$
$A=120^\circ $ atau $A=240^\circ $
Sudut A terkecil : sudut A terbesar
= $120^\circ :240^\circ $
= 1 : 2
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 6

Himpunan penyelesaian persamaan $\cos 2x-\sin x=0$ untuk $0\le x\le 2\pi $ adalah ….
A. $\left\{ \frac{\pi }{2},\frac{\pi }{3},\frac{\pi }{6} \right\}$
B. $\left\{ \frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2} \right\}$
C. $\left\{ \frac{\pi }{2},\frac{\pi }{6},\frac{7\pi }{6} \right\}$
D. $\left\{ \frac{7\pi }{6},\frac{4\pi }{3},\frac{11\pi }{6} \right\}$
E. $\left\{ \frac{4\pi }{3},\frac{11\pi }{6},2\pi \right\}$

Penyelesaian: Lihat/Tutup

Perbandingan Trigonometri Sudut Berelasi, $\sin \alpha =\cos (90^\circ -\alpha )$.
$\begin{align}\cos 2x-\sin x &= 0 \\ \cos 2x &= \sin x \\ \cos 2x &= \cos (90^\circ -x) \end{align}$
Persamaan trigonometri dasar,$\cos f(x)=\cos g(x)$ dengan $f(x)=2x$ dan $g(x)=90^\circ -x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x &= 90^\circ -x+k.360^\circ \\ 2x+x &= 90^\circ +k.360^\circ \\ 3x &= 90^\circ +k.360^\circ \\ x &= 30^\circ +k.120^\circ \end{align}$
$k=0\to x=30^\circ $
$k=1\to x=150^\circ $
$k=2\to x=270^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}2x &= -(90^\circ -x)+k.360^\circ \\ 2x &= -90^\circ +x+k.360^\circ \\ 2x-x &= -90^\circ +k.360^\circ \\ x &= -90^\circ +k.360^\circ \end{align}$
$k=1\to x=270^\circ $
HP = $\left\{ 30^\circ ,150^\circ ,270^\circ \right\}$ atau
HP = $\left\{ \frac{30^\circ }{180^\circ }\pi ,\frac{150^\circ }{180^\circ }\pi ,\frac{270^\circ }{180^\circ }\pi \right\}$ = $\left\{ \frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2} \right\}$
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 7

Himpunan penyelesaian persamaan $\cos 2x+\cos x=0$ untuk $0^\circ \le x\le 180^\circ $ adalah ….
A. $\{45^\circ ,120^\circ \}$
B. $\{45^\circ ,135^\circ \}$
C. $\{60^\circ ,135^\circ \}$
D. $\{60^\circ ,120^\circ \}$
E. $\{60^\circ ,180^\circ \}$

Penyelesaian: Lihat/Tutup

Perbandingan Trigonometri Sudut Berelasi, $-\cos \alpha =\cos (180^\circ -\alpha )$.
$\begin{align}\cos 2x+\cos x &= 0 \\ \cos 2x &= -\cos x \\ \cos 2x &= \cos (180^\circ +x) \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=2x$ dan $g(x)=180^\circ +x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x &= (180^\circ +x)+k.360^\circ \\ 2x-x &= 180^\circ +k.360^\circ \\ x &= 180^\circ +k.360^\circ \end{align}$
$k=0\to x=180^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}2x &= -(180^\circ +x)+k.360^\circ \\ 2x &= -180^\circ -x+k.360^\circ \\ 2x+x &= -180^\circ +k.360^\circ \\ 3x &= -180^\circ +k.360^\circ \\ x &= -60^\circ +k.120^\circ \end{align}$
$k=1\to x=60^\circ $
HP = $\{60^\circ ,180^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 8

Himpunan penyelesaian persamaan $\cos (2x+60^\circ )=-\frac{1}{2}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,90^\circ ,210^\circ ,270^\circ \}$
B. $\{60^\circ ,120^\circ ,210^\circ ,270^\circ \}$
C. $\{30^\circ ,120^\circ ,210^\circ ,300^\circ \}$
D. $\{60^\circ ,90^\circ ,240^\circ ,270^\circ \}$
E. $\{30^\circ ,90^\circ ,240^\circ ,300^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\cos (2x+60^\circ ) &= -\frac{1}{2} \\ \cos (2x+60^\circ ) &= \cos 120^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=2x+60^\circ $ dan $g(x)=120^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x+60^\circ &= 120^\circ +k.360^\circ \\ 2x &= 120^\circ -60^\circ +k.360^\circ \\ 2x &= 60^\circ +k.360^\circ \\ x &= 30^\circ +k.180^\circ \end{align}$
$k=0\to x=30^\circ $
$k=1\to x=210^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}2x+60^\circ &= -120^\circ +k.360^\circ \\ 2x &= -120^\circ -60^\circ +k.360^\circ \\ 2x &= -180^\circ +k.360^\circ \\ x &= -90^\circ -k.180^\circ \end{align}$
$k=-1\to x=90^\circ $
$k=-2\to x=270^\circ $
HP = $\{30^\circ ,90^\circ ,210^\circ ,270^\circ \}$
Jawaban: A

Soal Persamaan Trigonometri Dasar No. 9

Himpunan penyelesaian persamaan $1+2\sin x=0$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{120^\circ ,180^\circ \}$
B. $\{150^\circ ,260^\circ \}$
C. $\{180^\circ ,270^\circ \}$
D. $\{200^\circ ,320^\circ \}$
E. $\{210^\circ ,330^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}1+2\sin x &= 0 \\ 2\sin x &= -1 \\ \sin x &= -\frac{1}{2} \end{align}$
$x=210^\circ $ atau $x=330^\circ $
HP = $\{210^\circ ,330^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 10

Himpunan penyelesaian persamaan $2\cos x=1$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,120^\circ \}$
B. $\{30^\circ ,300^\circ \}$
C. $\{30^\circ ,330^\circ \}$
D. $\{60^\circ ,120^\circ \}$
E. $\{60^\circ ,300^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}2\cos x &= 1 \\ \cos x &= \frac{1}{2} \end{align}$
$x=60^\circ $ atau $x=300^\circ $
HP = $\{60^\circ ,300^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 11

Himpunan penyelesaian dari $\sin 3x=\frac{1}{2}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{10^\circ ,50^\circ ,130^\circ ,170^\circ ,250^\circ ,290^\circ \}$
B. $\{10^\circ ,50^\circ ,160^\circ ,170^\circ ,250^\circ ,290^\circ \}$
C. $\{10^\circ ,50^\circ ,160^\circ ,170^\circ ,250^\circ ,290^\circ \}$
D. $\{10^\circ ,60^\circ ,130^\circ ,170^\circ ,250^\circ ,290^\circ \}$
E. $\{10^\circ ,50^\circ ,130^\circ ,170^\circ ,250^\circ ,340^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sin 3x &= \frac{1}{2} \\ \sin 3x &= \sin 30^\circ \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=3x$ dan $g(x)=30^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}3x &= 30^\circ +k.360^\circ \\ x &= 10^\circ +k.120^\circ \end{align}$
$k=0\to x=10^\circ $
$k=1\to x=130^\circ $
$k=2\to x=250^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}3x &= 180^\circ -30^\circ +k.360^\circ \\ 3x &= 150^\circ +k.360^\circ \\ x &= 50^\circ +k.120^\circ \end{align}$
$k=0\to x=50^\circ $
$k=1\to x=170^\circ $
$k=2\to x=290^\circ $
HP = $\{10^\circ ,50^\circ ,130^\circ ,170^\circ ,250^\circ ,290^\circ \}$
Jawaban: A

Soal Persamaan Trigonometri Dasar No. 12

Himpunan penyelesaian dari $\cos 5x=\frac{1}{2}\sqrt{2}$ untuk $0^\circ \le x\le 180^\circ $ adalah ….
A. $\{10^\circ ,63^\circ ,81^\circ ,135^\circ ,153^\circ \}$
B. $\{9^\circ ,63^\circ ,91^\circ ,135^\circ \}$
C. $\{9^\circ ,63^\circ ,81^\circ ,135^\circ ,153^\circ \}$
D. $\{9^\circ ,73^\circ ,81^\circ ,153^\circ \}$
E. $\{9^\circ ,83^\circ ,135^\circ ,153^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\cos 5x &= \frac{1}{2}\sqrt{2} \\ \cos 5x &= \cos 45^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=5x$ dan $\cos g(x)$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}5x &= 45^\circ +k.360^\circ \\ x &= 9^\circ +k.72^\circ \end{align}$
$k=0\to x=9^\circ $
$k=1\to x=81^\circ $
$k=2\to x=153^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}5x &= -45^\circ +k.360^\circ \\ x &= -9^\circ +k.72^\circ \end{align}$
$k=1\to x=63^\circ $
$k=2\to x=135^\circ $
HP = $\{9^\circ ,63^\circ ,81^\circ ,135^\circ ,153^\circ \}$
Jawaban: C

Soal Persamaan Trigonometri Dasar No. 13

Himpunan penyelesaian dari persamaan $\tan 4x=\sqrt{3}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{15^\circ ,60^\circ ,145^\circ ,150^\circ ,195^\circ ,240^\circ ,285^\circ ,330^\circ \}$
B. $\{15^\circ ,60^\circ ,105^\circ ,150^\circ ,185^\circ ,240^\circ ,285^\circ ,330^\circ \}$
C. $\{25^\circ ,60^\circ ,105^\circ ,150^\circ ,195^\circ ,240^\circ ,285^\circ ,330^\circ \}$
D. $\{15^\circ ,60^\circ ,105^\circ ,150^\circ ,195^\circ ,240^\circ ,285^\circ ,330^\circ \}$
E. $\{15^\circ ,60^\circ ,105^\circ ,150^\circ ,195^\circ ,240^\circ ,285^\circ ,340^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\tan 4x &= \sqrt{3} \\ \tan 4x &= \tan 60^\circ \end{align}$
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=4x$ dan $g(x)=60^\circ $ maka:
$\begin{align}\color{blue}f(x) &\color{blue}= g(x)+k.180^\circ \\ 4x &= 60^\circ +k.180^\circ \\ x &= 15^\circ +k.45^\circ \end{align}$
$x=0\to x=15^\circ $
$x=1\to x=60^\circ $
$x=2\to x=105^\circ $
$x=3\to x=150^\circ $
$x=4\to x=195^\circ $
$x=5\to x=240^\circ $
$x=6\to x=285^\circ $
$x=7\to x=330^\circ $
HP = $\{15^\circ ,60^\circ ,105^\circ ,150^\circ ,195^\circ ,240^\circ ,285^\circ ,330^\circ \}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 14

Jika $\tan 2x=\tan \frac{\pi }{3}$, maka harga $x$ adalah ….
A. $\left\{ \frac{\pi }{4},\frac{5\pi }{4},\frac{2\pi }{3},\frac{4\pi }{3} \right\}$
B. $\left\{ \frac{\pi }{3},\frac{5\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3} \right\}$
C. $x=\frac{\pi }{2}+\frac{1}{2}k\pi $
D. $\left\{ 0,\frac{\pi }{2},\pi \right\}$
E. $x=\frac{\pi }{6}+\frac{1}{2}k\pi $

Penyelesaian: Lihat/Tutup

$\tan 2x=\tan \frac{\pi }{3}$
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=2x$ dan $g(x)=\frac{\pi }{3}$ maka:
$\begin{align}f(x) &= g(x)+k.\pi \\ 2x &= \frac{\pi }{3}+k.\pi \\ x &= \frac{\pi }{6}+\frac{1}{2}k\pi \end{align}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 15

Himpunan penyelesaian dari $\sin (x-30^\circ )=\frac{1}{2}\sqrt{3}$ untuk $0^\circ \le x\le 720^\circ $ adalah ….
A. $\{90^\circ ,150^\circ ,450^\circ ,510^\circ \}$
B. $\{30^\circ ,90^\circ ,150^\circ ,300^\circ \}$
C. $\{60^\circ ,90^\circ ,300^\circ ,450^\circ ,510^\circ \}$
D. $\{90^\circ ,150^\circ ,300^\circ ,450^\circ \}$
E. $\{30^\circ ,60^\circ ,90^\circ ,150^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sin (x-30^\circ ) &= \frac{1}{2}\sqrt{3} \\ \sin (x-30^\circ ) &= \sin 60^\circ \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=x-30^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-30^\circ &= 60^\circ +k.360^\circ \\ x &= 60^\circ +30^\circ +k.360^\circ \\ x &= 90^\circ +k.360^\circ \end{align}$
$k=0\to x=90^\circ $
$k=1\to x=450^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}x-30^\circ &= (180^\circ -60^\circ )+k.360^\circ \\ x-30^\circ &= 120^\circ +k.360^\circ \\ x &= 120^\circ +30^\circ +k.360^\circ \\ x &= 150^\circ +k.360^\circ \end{align}$
$k=0\to x=150^\circ $
$k=1\to x=510^\circ $
HP = $\{90^\circ ,150^\circ ,450^\circ ,510^\circ \}$
Jawaban: A

Soal Persamaan Trigonometri Dasar No. 16

Himpunan penyelesaian dari persamaan $2\cos 3x=1$ untuk $0^\circ \le x\le 180^\circ $ adalah ….
A. $\{0^\circ ,20^\circ ,60^\circ \}$
B. $\{0^\circ ,20^\circ ,100^\circ \}$
C. $\{20^\circ ,60^\circ ,100^\circ \}$
D. $\{20^\circ ,100^\circ ,140^\circ \}$
E. $\{100^\circ ,140^\circ ,180^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}2\cos 3x &= 1 \\ \cos 3x &= \frac{1}{2} \\ \cos 3x &= \cos 60^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=3x$ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}3x &= 60^\circ +k.360^\circ \\ x &= 20^\circ +k.120^\circ \end{align}$
$k=0\to x=20^\circ $
$k=1\to x=140^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}3x &= -60^\circ +k.360^\circ \\ x &= -20^\circ +k.120^\circ \end{align}$
$k=1\to x=100^\circ $
HP = $\{20^\circ ,100^\circ ,140^\circ \}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 17

Himpunan penyelesaian persamaan $\cos (3x-45^\circ )=-\frac{1}{2}\sqrt{2}$ untuk $0^\circ \le x\le 180^\circ $ adalah ….
A. $\{60^\circ ,90^\circ ,180^\circ \}$
B. $\{60^\circ ,90^\circ \}$
C. $\{90^\circ ,180^\circ \}$
D. $\{60^\circ ,120^\circ \}$
E. $\{30^\circ ,150^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\cos (3x-45^\circ ) &= -\frac{1}{2}\sqrt{2} \\ \cos (3x-45^\circ ) &= \cos 135^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=3x-45^\circ $ dan $g(x)=135^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}3x-45^\circ &= 135^\circ +k.360^\circ \\ 3x &= 135^\circ +45^\circ +k.360^\circ \\ 3x &= 180^\circ +k.360^\circ \\ x &= 60^\circ +k.120^\circ \end{align}$
$k=0\to x=60^\circ $
$k=1\to x=180^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}3x-45^\circ &= -135^\circ +k.360^\circ \\ 3x &= -135^\circ +45^\circ +k.360^\circ \\ 3x &= -90^\circ +k.360^\circ \\ x &= -30^\circ +k.120^\circ \end{align}$
$k=1\to x=90^\circ $
HP = $\{60^\circ ,90^\circ ,180^\circ \}$
Jawaban: A

Soal Persamaan Trigonometri Dasar No. 18

Diketahui persamaan $\sqrt{2}\cos x-1=0$, $0^\circ \le x\le 180^\circ $. Himpunan penyelesaian persamaan tersebut adalah ….
A. $\{45^\circ ,135^\circ \}$
B. $\{90^\circ ,270^\circ \}$
C. $\{45^\circ ,315^\circ \}$
D. $\{45^\circ \}$
E. $\{45^\circ ,135^\circ ,225^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{2}\cos x-1 &= 0 \\ \sqrt{2}\cos x &= 1 \\ \cos x &= \frac{1}{\sqrt{2}} \\ \cos x &= \frac{1}{2}\sqrt{2} \\ x &= 45^\circ \end{align}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 19

Himpunan penyelesaian dari $\cos (x-30^\circ )=\frac{1}{2}\sqrt{2}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{75^\circ ,300^\circ \}$
B. $\{75^\circ ,345^\circ \}$
C. $\{50^\circ ,250^\circ \}$
D. $\{65^\circ ,345^\circ \}$
E. $\{60^\circ ,250^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\cos (x-30^\circ ) &= \frac{1}{2}\sqrt{2} \\ \cos (x-30^\circ ) &= \cos 45^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-30^\circ $ dan $g(x)=45^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-30^\circ &= 45^\circ +k.360^\circ \\ x &= 45^\circ +30^\circ +k.360^\circ \\ x &= 75^\circ +k.360^\circ \end{align}$
$k=0\to x=75^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-30^\circ &= -45^\circ +k.360^\circ \\ x &= -45^\circ +30^\circ +k.360^\circ \\ x &= -15^\circ +k.360^\circ \end{align}$
$k=1\to x=345^\circ $
HP = $\{75^\circ ,345^\circ \}$
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 20

Untuk $0^\circ \le x\le 180^\circ $, himpunan penyelesaian persamaan trigonometri $4\sin x-2=0$ adalah ….
A. $\{30^\circ \}$
B. $\{30^\circ ,150^\circ \}$
C. $\{60^\circ \}$
D. $\{60^\circ ,120^\circ \}$
E. $\{45^\circ ,145^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}4\sin x-2 &= 0 \\ 4\sin x &= 2 \\ \sin x &= \frac{2}{4} \\ \sin x &= \frac{1}{2} \end{align}$
$x=30^\circ $ atau $x=150^\circ $
HP = $\{30^\circ ,150^\circ \}$
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 21

Himpunan penyelesaian dari persamaan $\sin x=\frac{1}{2}\sqrt{3}$ untuk $0\le x\le 2\pi $ adalah ….
A. $\left\{ \frac{\pi }{3},\frac{2\pi }{3} \right\}$
B. $\left\{ \frac{\pi }{3},\frac{\pi }{6} \right\}$
C. $\left\{ \frac{\pi }{3},\frac{\pi }{2} \right\}$
D. $\left\{ \frac{\pi }{3},\frac{5\pi }{6} \right\}$
E. $\left\{ \frac{2\pi }{3},\frac{5\pi }{6} \right\}$

Penyelesaian: Lihat/Tutup

$\sin x=\frac{1}{2}\sqrt{3}$
$x=60^\circ =\frac{60^\circ }{180^\circ }\pi =\frac{\pi }{3}$ atau $x=120^\circ =\frac{120^\circ }{180^\circ }\pi =\frac{2\pi }{3}$
HP = $\left\{ \frac{\pi }{3},\frac{2\pi }{3} \right\}$
Jawaban: A

Soal Persamaan Trigonometri Dasar No. 22

Himpunan penyelesaian dari persamaan trigonometri $\cos 2x+\cos x=0$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,60^\circ ,180^\circ \}$
B. $\{30^\circ ,90^\circ ,150^\circ \}$
C. $\{30^\circ ,180^\circ ,300^\circ \}$
D. $\{60^\circ ,120^\circ ,270^\circ \}$
E. $\{60^\circ ,180^\circ ,300^\circ \}$

Penyelesaian: Lihat/Tutup

Perbandingan trigonometri sudut berelasi, $-\cos \alpha =\cos (180^\circ -\alpha )$.
$\begin{align}\cos 2x+\cos x &= 0 \\ \cos 2x &= -\cos x \\ \cos 2x &= \cos (180^\circ -x) \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=2x$ dan $g(x)=180^\circ -x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x &= 180^\circ -x+k.360^\circ \\ 2x+x &= 180^\circ +k.360^\circ \\ 3x &= 180^\circ +k.360^\circ \\ x &= 60^\circ +k.120^\circ \end{align}$
$k=0\to x=60^\circ $
$k=1\to x=180^\circ $
$k=2\to x=300^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}2x &= -(180^\circ -x)+k.360^\circ \\ 2x &= -180^\circ +x+k.360^\circ \\ 2x-x &= -180^\circ +k.360^\circ \\ x &= -180^\circ +k.360^\circ \end{align}$
$k=1\to x=180^\circ $
HP = $\{60^\circ ,180^\circ ,300^\circ \}$
Jawaban: E

Soal Persamaan Trigonometri Dasar No. 23

Himpunan penyelesaian dari persamaan $\sin (x-60^\circ )=\cos 2x$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{70^\circ ,170^\circ ,210^\circ ,250^\circ \}$
B. $\{70^\circ ,190^\circ ,210^\circ ,250^\circ \}$
C. $\{50^\circ ,190^\circ ,250^\circ ,290^\circ \}$
D. $\{50^\circ ,170^\circ ,210^\circ ,290^\circ \}$
E. $\{50^\circ ,170^\circ ,250^\circ ,290^\circ \}$

Penyelesaian: Lihat/Tutup

Perbandingan trigonometri sudut berelasi, $\sin (90^\circ -\alpha )=\cos \alpha $
$\begin{align}\sin (x-60^\circ ) &= \cos 2x \\ \sin (x-60^\circ ) &= \sin (90^\circ -2x) \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=x-60^\circ $ dan $g(x)=90^\circ -2x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-60^\circ &= 90^\circ -2x+k.360^\circ \\ x+2x &= 90^\circ +60^\circ +k.360^\circ \\ 3x &= 150^\circ +k.360^\circ \\ x &= 50^\circ +k.120^\circ \end{align}$
$k=0\to x=50^\circ $
$k=1\to x=170^\circ $
$k=2\to x=290^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}x-60^\circ &= (180^\circ -(90^\circ -2x))+k.360^\circ \\ x-60^\circ &= 90^\circ +2x+k.360^\circ \\ x-2x &= 90^\circ +60^\circ +k.360^\circ \\ -x &= 150^\circ +k.360^\circ \\ x &= -150^\circ -k.360^\circ \end{align}$
$k=-1\to x=210^\circ $
HP = $\{50^\circ ,170^\circ ,210^\circ ,290^\circ \}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 24

Himpunan penyelesaian dari persamaan $\sqrt{6}\tan 2x-\sqrt{2}=0$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{15^\circ ,105^\circ ,195^\circ ,315^\circ \}$
B. $\{15^\circ ,195^\circ ,225^\circ ,315^\circ \}$
C. $\{15^\circ ,105^\circ ,195^\circ ,285^\circ \}$
D. $\{105^\circ ,195^\circ ,255^\circ ,315^\circ \}$
E. $\{105^\circ ,185^\circ ,255^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{6}\tan 2x-\sqrt{2} &= 0 \\ \sqrt{6}\tan 2x &= \sqrt{2} \\ \tan 2x &= \frac{\sqrt{2}}{\sqrt{6}} \\ \tan 2x &= \frac{1}{\sqrt{3}} \\ \tan 2x &= \frac{1}{3}\sqrt{3} \\ \tan 2x &= \tan 30^\circ \end{align}$
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=2x$ dan $g(x)=30^\circ $ maka:
$\begin{align}\color{blue}f(x) &\color{blue}= g(x)+k.180^\circ \\ 2x &= 30^\circ +k.180^\circ \\ x &= 15^\circ +k.90^\circ \end{align}$
$k=0\to x=15^\circ $
$k=1\to x=105^\circ $
$k=2\to x=195^\circ $
$k=3\to x=285^\circ $
HP = $\{15^\circ ,105^\circ ,195^\circ ,285^\circ \}$
Jawaban: C

Soal Persamaan Trigonometri Dasar No. 25

Penyelesaian dari $\cos (40^\circ +x)+\sin (40^\circ +x)=0$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $x=45^\circ $ atau $x=135^\circ $
B. $x=-95^\circ $ atau $x=275^\circ $
C. $x=95^\circ $ atau $x=275^\circ $
D. $x=5^\circ $ atau $x=95^\circ $
E. $x=85^\circ $ atau $x=5^\circ $

Penyelesaian: Lihat/Tutup

Perbandingan trigonometri sudut berelasi, $-\sin \alpha =\cos (90^\circ +\alpha )$ maka:
$\cos (40^\circ +x)+\sin (40^\circ +x)=0$
$\begin{align}\cos (40^\circ +x) &= -\sin (40^\circ +x) \\ \cos (40^\circ +x) &= \cos (90^\circ +(40^\circ +x)) \\ \cos (40^\circ +x) &= \cos (130^\circ +x) \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=40^\circ +x$ dan $g(x)=130^\circ +x$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}40^\circ +x &= 130^\circ +x+k.360^\circ \\ x-x &= 130^\circ -40^\circ +k.360^\circ \\ 0 &= 90^\circ +k.360^\circ \end{align}$
(tidak ada penyelesaian)
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}40^\circ +x &= -(130^\circ +x)+k.360^\circ \\ 40^\circ +x &= -130^\circ -x+k.360^\circ \\ x+x &= -130^\circ -40^\circ +k.360^\circ \\ 2x &= -170^\circ +k.360^\circ \\ x &= -85^\circ +k.180^\circ \end{align}$
$k=1\to x=95^\circ $
$k=2\to x=275^\circ $
Jadi, $x=95^\circ $ atau $x=275^\circ $
Jawaban: C

Soal Persamaan Trigonometri Dasar No. 26

Himpunan penyelesaian dari $6\sin (2x+60^\circ )=3$ untuk $0^\circ \le x\le 180^\circ $ adalah ….
A. $\{30^\circ ,150^\circ \}$
B. $\{45^\circ ,165^\circ \}$
C. $\{15^\circ ,150^\circ \}$
D. $\{30^\circ ,60^\circ \}$
E. $\{120^\circ ,135^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}6\sin (2x+60^\circ ) &= 3 \\ \sin (2x+60^\circ ) &= \frac{3}{6} \\ \sin (2x+60^\circ ) &= \frac{1}{2} \\ \sin (2x+60^\circ ) &= \sin 30^\circ \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=2x+60^\circ $ dan $g(x)=30^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x+60^\circ &= 30^\circ +k.360^\circ \\ 2x &= 30^\circ -60^\circ +k.360^\circ \\ 2x &= -30^\circ +k.360^\circ \\ x &= -15^\circ +k.180^\circ \end{align}$
$k=1\to x=165^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x+60^\circ &= (180^\circ -30^\circ )+k.360^\circ \\ 2x+60^\circ &= 150^\circ +k.360^\circ \\ 2x &= 150^\circ -60^\circ +k.360^\circ \\ 2x &= 90^\circ +k.360^\circ \\ x &= 45^\circ +k.180^\circ \end{align}$
$k=0\to x=45^\circ $
HP = $\{45^\circ ,165^\circ \}$
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 27

Himpunan penyelesaian dari $\sin (x-75^\circ )=\frac{1}{2}\sqrt{3}$ dengan $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{60^\circ ,135^\circ \}$
B. $\{60^\circ ,195^\circ \}$
C. $\{135^\circ ,195^\circ \}$
D. $\{135^\circ ,315^\circ \}$
E. $\{195^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sin (x-75^\circ ) &= \frac{1}{2}\sqrt{3} \\ \sin (x-75^\circ ) &= \sin 60^\circ \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=x-75^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-75^\circ &= 60^\circ +k.360^\circ \\ x &= 60^\circ +75^\circ +k.360^\circ \\ x &= 135^\circ +k.360^\circ \end{align}$
$k=0\to x=135^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}x-75^\circ &= (180^\circ -60^\circ )+k.360^\circ \\ x-75^\circ &= 120^\circ +k.360^\circ \\ x &= 120^\circ +75^\circ +k.360^\circ \\ x &= 195^\circ +k.360^\circ \end{align}$
$k=0\to x=195^\circ $
HP = $\{135^\circ ,195^\circ \}$
Jawaban: C

Soal Persamaan Trigonometri Dasar No. 28

Nilai $x$ yang memenuhi persamaan trigonometri $\sqrt{3}+3\tan (2x-30^\circ )=0$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{60^\circ ,180^\circ ,240^\circ ,360^\circ \}$
B. $\{90^\circ ,180^\circ ,270^\circ ,360^\circ \}$
C. $\{60^\circ ,150^\circ ,270^\circ ,330^\circ \}$
D. $\{90^\circ ,150^\circ ,210^\circ ,360^\circ \}$
E. $\{90^\circ ,120^\circ ,270^\circ ,330^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{3}+3\tan (2x-30^\circ ) &= 0 \\ 3\tan (2x-30^\circ ) &= -\sqrt{3}\\ \tan (2x-30^\circ ) &= -\frac{1}{3}\sqrt{3} \\ \tan (2x-30^\circ ) &= \tan 150^\circ \end{align}$
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=2x-30^\circ $ dan $g(x)=150^\circ $ maka:
$\begin{align}\color{blue}f(x) &\color{blue}= g(x)+k.180^\circ \\ 2x-30^\circ &= 150^\circ +k.180^\circ \\ 2x &= 150^\circ +30^\circ +k.180^\circ \\ 2x &= 180^\circ +k.180^\circ \\ x &= 90^\circ +k.90^\circ \end{align}$
$k=0\to x=90^\circ $
$k=1\to x=180^\circ $
$k=2\to x=270^\circ $
$k=3\to x=360^\circ $
HP = $\{90^\circ ,180^\circ ,270^\circ ,360^\circ \}$
Jawaban: B

Soal Persamaan Trigonometri Dasar No. 29

Nilai $x$ yang memenuhi persamaan trigonometri $2+\sqrt{12}\sin (2x+30^\circ )=5$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{45^\circ ,135^\circ ,195^\circ ,225^\circ \}$
B. $\{15^\circ ,75^\circ ,195^\circ ,245^\circ \}$
C. $\{45^\circ ,75^\circ ,195^\circ ,225^\circ \}$
D. $\{15^\circ ,45^\circ ,195^\circ ,225^\circ \}$
E. $\{15^\circ ,45^\circ ,135^\circ ,3155^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}2+\sqrt{12}\sin (2x+30^\circ ) &= 5 \\ \sqrt{12}\sin (2x+30^\circ ) &= 5-2 \\ 2\sqrt{3}\sin (2x+30^\circ ) &= 3 \\ \sin (2x+30^\circ ) &= \frac{3}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \sin (2x+30^\circ ) &= \frac{1}{2}\sqrt{3} \\ \sin (2x+30^\circ ) &= \sin 60^\circ \end{align}$
Persamaan trigonometri dasar, $\sin f(x)=\sin g(x)$ dengan $f(x)=2x+30^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x+30^\circ &= 60^\circ +k.360^\circ \\ 2x &= 60^\circ -30^\circ +k.360^\circ \\ 2x &= 30^\circ +k.360^\circ \\ x &= 15^\circ +k.180^\circ \end{align}$
$k=0\to x=15^\circ $
$k=1\to x=195^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x+30^\circ &= (180^\circ -60^\circ )+k.360^\circ \\ 2x+30^\circ &= 120^\circ +k.360^\circ \\ 2x &= 120^\circ -30^\circ +k.360^\circ \\ 2x &= 90^\circ +k.360^\circ \\ x &= 45^\circ +k.180^\circ \end{align}$
$k=0\to x=45^\circ $
$k=1\to x=225^\circ $
HP = $\{15^\circ ,45^\circ ,195^\circ ,225^\circ \}$
Jawaban: D

Soal Persamaan Trigonometri Dasar No. 30

Himpunan penyelesaian dari persamaan $\tan (2x-30^\circ )=-\sqrt{3}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{75^\circ ,165^\circ ,255^\circ ,345^\circ \}$
B. $\{105^\circ ,185^\circ ,255^\circ ,315^\circ \}$
C. $\{75^\circ ,105^\circ ,165^\circ ,205^\circ \}$
D. $\{75^\circ ,165^\circ ,225^\circ ,315^\circ \}$
E. $\{75^\circ ,165^\circ ,255^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}\tan (2x-30^\circ ) &= -\sqrt{3} \\ \tan (2x-30^\circ ) &= \tan 120^\circ \end{align}$
Persamaan trigonometri dasar, $\tan f(x)=\tan g(x)$ dengan $f(x)=2x-30^\circ $ dan $g(x)=120^\circ $ maka:
$\begin{align}\color{blue}f(x) &\color{blue}= g(x)+k.180^\circ \\ 2x-30^\circ &= 120^\circ +k.180^\circ \\ 2x &= 120^\circ +30^\circ +k.180^\circ \\ 2x &= 150^\circ +k.180^\circ \\ x &= 75^\circ +k.90^\circ \end{align}$
$k=0\to x=75^\circ $
$k=1\to x=165^\circ $
$k=2\to x=255^\circ $
$k=3\to x=345^\circ $
HP = $\{75^\circ ,165^\circ ,255^\circ ,345^\circ \}$
Jawaban: A

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