Soal Persamaan Trigonometri a cos x + b sin x = c dan Pembahasan

Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) berbagi kumpulan soal Persamaan Trigonometri a cos x + b sin x = c berserta pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.

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klik "LIHAT/TUTUP:".

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Soal No. 1

Diketahui persamaan $2{{\cos }^2}x+\sqrt{3}\sin 2x=1+\sqrt{3}$ untuk $0 < x < \frac{\pi }{2}$. Nilai $x$ yang memenuhi adalah ….
A. $\frac{\pi }{6}$ dan $\frac{\pi }{2}$
B. $\frac{\pi }{3}$ dan $\frac{5\pi }{12}$
C. $\frac{\pi }{12}$ dan $\frac{5\pi }{12}$
D. $\frac{\pi }{12}$ dan $\frac{\pi }{4}$
E. $\frac{\pi }{6}$ dan $\frac{\pi }{4}$

Penyelesaian: Lihat/Tutup

Rumus Trigonometri Sudut Ganda, $2{{\cos }^2}\alpha -1=\cos 2\alpha $.
$\begin{align}2{{\cos }^2}x+\sqrt{3}\sin 2x &= 1+\sqrt{3} \\ 2{{\cos }^2}x-1+\sqrt{3}\sin 2x &=\sqrt{3} \\ \cos 2x+\sqrt{3}\sin 2x &= \sqrt{3} \end{align}$
$a=1$ (positif) dan $b=\sqrt{3}$ (positif) maka $\theta $ di kuadran I
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{\sqrt{3}}{1} \right) \\ &= {\tan }^{-1}\left( \sqrt{3} \right) \\ \theta &= 60^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{1^2+(\sqrt{3})^2} \\ k &= 2 \end{align}$
$\begin{align}\cos 2x+\sqrt{3}\sin 2x &= \sqrt{3} \\ k\cos (2x-\theta ) &= c \\ 2\cos (2x-60^\circ ) &= \sqrt{3} \\ \cos (2x-60^\circ ) &= \frac{1}{2}\sqrt{3} \\ \cos (2x-60^\circ ) &= \cos 30^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=2x-60^\circ $ dan $g(x)=30^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x-60^\circ &= 30^\circ +k.360^\circ \\ 2x &= 30^\circ +60^\circ +k.360^\circ \\ 2x &= 90^\circ +k.360^\circ \\ x &= 45^\circ +k.180^\circ \end{align}$
$k=0\to x=45^\circ $
2) $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=0\to x=15^\circ $
HP = $\{15^\circ ,45^\circ \}$ atau
HP = $\left\{ \frac{15^\circ }{180^\circ }\pi ,\frac{45^\circ }{180^\circ }\pi \right\}$ = $\left\{ \frac{\pi }{12},\frac{\pi }{4} \right\}$
Jawaban: D

Soal No. 2

Nilai $x$ yang memenuhi persamaan $2\cos x+2\sin x=\sqrt{2}$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $15^\circ $ atau $135^\circ $
B. $45^\circ $ atau $315^\circ $
C. $75^\circ $ atau $375^\circ $
D. $105^\circ $ atau $345^\circ $
E. $165^\circ $ atau $285^\circ $

Penyelesaian: Lihat/Tutup

$2\cos x+2\sin x=\sqrt{2}$
$a=2$ (positif) dan $b=2$ (positif) maka $\theta $ di kuadran I
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{2}{2} \right) \\ &= {\tan }^{-1}1 \\ \theta &= 45^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{2^2+2^2} \\ &= \sqrt{8} \\ k &= 2\sqrt{2} \end{align}$
$\begin{align}2\cos x+2\sin x &= \sqrt{2} \\ k\cos (x-\theta ) &= c \\ 2\sqrt{2}\cos (x-45^\circ ) &= \sqrt{2} \\ \cos (x-45^\circ ) &= \frac{\sqrt{2}}{2\sqrt{2}} \\ \cos (x-45^\circ ) &= \frac{1}{2} \\ \cos (x-45^\circ ) &= \cos 60^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-45^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-45^\circ &= 60^\circ +k.360^\circ \\ x &= 60^\circ +45^\circ +k.360^\circ \\ x &= 105^\circ +k.360^\circ \end{align}$
$k=0\to x=105^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-45^\circ &= -60^\circ +k.360^\circ \\ x &= -60^\circ +45^\circ +k.360^\circ \\ x &= -15^\circ +k.360^\circ \end{align}$
$k=1\to x=345^\circ $
Jadi, nilai $x$ yang memenuhi adalah $105^\circ $ atau $345^\circ $.
Jawaban: D

Soal No. 3

Nilai $x$ yang memenuhi $\sqrt{3}\cos x+\sin x=\sqrt{2}$ untuk $0\le x\le 2\pi $ adalah ….
A. $\frac{1}{12}\pi $ atau $\frac{11}{12}\pi $
B. $\frac{1}{12}\pi $ atau $\frac{23}{12}\pi $
C. $\frac{5}{12}\pi $ atau $\frac{7}{12}\pi $
D. $\frac{5}{12}\pi $ atau $\frac{19}{12}\pi $
E. $\frac{5}{12}\pi $ atau $\frac{23}{12}\pi $

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x+\sin x=\sqrt{2}$
$a=\sqrt{3}$ (positif) dan $b=1$ (positif) maka $\theta $ di kuadran I.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( \frac{1}{3}\sqrt{3} \right) \\ \theta &= 30^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{3})^2+1^2} \\ k &= 2 \end{align}$
$\begin{align}\sqrt{3}\cos x+\sin x &= \sqrt{2} \\ k\cos (x-\theta ) &= \sqrt{2} \\ 2\cos (x-30^\circ ) &= \sqrt{2} \\ \cos (x-30^\circ ) &= \frac{1}{2}\sqrt{2} \\ \cos (x-30^\circ ) &= 45^\circ \end{align}$
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 = $\{75^\circ ,345^\circ \}$ atau HP = $\left\{ \frac{5}{12}\pi ,\frac{23}{12}\pi \right\}$
Jawaban: E

Soal No. 4

Untuk $0^\circ \le x\le 360^\circ $, himpunan penyelesaian dari $\sin x-\sqrt{3}\cos x-\sqrt{3}=0$ adalah ….
A. $\{120^\circ ,180^\circ \}$
B. $\{90^\circ ,210^\circ \}$
C. $\{30^\circ ,270^\circ \}$
D. $\{0^\circ ,300^\circ \}$
E. $\{0^\circ ,300^\circ ,360^\circ \}$

Penyelesaian: Lihat/Tutup

$\sin x-\sqrt{3}\cos x-\sqrt{3}=0$
$-\sqrt{3}\cos x+\sin x=\sqrt{3}$
$a=-\sqrt{3}$ (negatif) dan $b=1$ (positif) maka $\theta $ di kuadran II.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{1}{-\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( -\frac{1}{3}\sqrt{3} \right) \\ \theta &= 150^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(-\sqrt{3})^2+1^2} \\ k &= 2 \end{align}$
$\begin{align}-\sqrt{3}\cos x+\sin x &= \sqrt{3} \\ k\cos (x-\theta ) &= c \\ 2\cos (x-150^\circ ) &= \sqrt{3} \\ \cos (x-150^\circ ) &= \frac{1}{2}\sqrt{3} \\ \cos (x-150^\circ ) &= \cos 30^\circ \end{align}$
Persamaan Trigonometri Dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-150^\circ $ dan $g(x)=30^\circ $.
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-150^\circ &= 30^\circ +k.360^\circ \\ x &= 30^\circ +150^\circ +k.360^\circ \\ x &= 180^\circ +k.360^\circ \end{align}$
$k=0\to x=180^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-150^\circ &= -30^\circ +k.360^\circ \\ x &= -30^\circ +150^\circ +k.360^\circ \\ x &= 120^\circ +k.360^\circ \end{align}$
$k=0\to x=120^\circ $
HP = $\{120^\circ ,180^\circ \}$
Jawaban: A

Soal No. 5

Himpunan penyelesaian persamaan $2\sqrt{3}\cos 2x-4\sin x\cos x=2$ untuk $0\le x\le 2\pi $ adalah ….
A. $\left\{ \frac{\pi }{4},\frac{5\pi }{6},\frac{13\pi }{12},\frac{13\pi }{6} \right\}$
B. $\left\{ \frac{\pi }{6},\frac{\pi }{2},\frac{13\pi }{12},\frac{7\pi }{6} \right\}$
C. $\left\{ \frac{\pi }{12},\frac{3\pi }{4},\frac{13\pi }{12},\frac{7\pi }{4} \right\}$
D. $\left\{ \frac{\pi }{6},\frac{3\pi }{4},\frac{13\pi }{12},\frac{7\pi }{4} \right\}$
E. $\left\{ \frac{3\pi }{4},\frac{5\pi }{6},\frac{13\pi }{12},\frac{5\pi }{3} \right\}$

Penyelesaian: Lihat/Tutup

$\begin{align}2\sqrt{3}\cos 2x-4\sin x\cos x &= 2 \\ 2\sqrt{3}\cos 2x-2.2\sin x\cos x &= 2 \\ 2\sqrt{3}\cos 2x-2\sin 2x &= 2 \end{align}$
$a=2\sqrt{3}$ (positif) dan $b=-2$ (negatif) maka $\theta $ di kuadran IV
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-2}{2\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( \frac{-1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( -\frac{1}{3}\sqrt{3} \right) \\ \theta &= 330^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(2\sqrt{3})^2+(-2)^2} \\ k &= 4 \end{align}$
$\begin{align}2\sqrt{3}\cos 2x-2\sin 2x &= 2 \\ k\cos (2x-\theta ) &= c \\ 4\cos (2x-330^\circ ) &= 2 \\ \cos (2x-330^\circ ) &= \frac{2}{4} \\ \cos (2x-330^\circ ) &= \frac{1}{2} \\ \cos (2x-330^\circ ) &= \cos 60^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=2x-330^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x-330^\circ &= 60^\circ +k.360^\circ \\ 2x &= 60^\circ +330^\circ +k.360^\circ \\ 2x &= 390^\circ +k.360^\circ \\ x &= 195^\circ +k.180^\circ \end{align}$
$k=-1\to x=15^\circ $
$k=0\to x=195^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}2x-330^\circ &= -60^\circ +k.360^\circ \\ 2x &= -60^\circ +330^\circ +k.360^\circ \\ 2x &= 270^\circ +k.360^\circ \\ x &= 135^\circ +k.180^\circ \end{align}$
$k=0\to x=135^\circ $
$k=1\to x=315^\circ $
HP = $\{15^\circ ,135^\circ ,195^\circ ,315^\circ \}$ atau
HP = $\left\{ \frac{15^\circ }{180^\circ }\pi ,\frac{135^\circ }{180^\circ }\pi ,\frac{195^\circ }{180^\circ }\pi ,\frac{315^\circ }{180^\circ }\pi \right\}$ = $\left\{ \frac{1}{12}\pi ,\frac{3}{4}\pi ,\frac{13}{12}\pi ,\frac{7}{4}\pi \right\}$
Jawaban: C

Soal No. 6

Himpunan penyelesaian persamaan $-\sin x+\cos x=1$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{0^\circ ,150^\circ ,180^\circ \}$
B. $\{0^\circ ,270^\circ ,360^\circ \}$
C. $\{30^\circ ,150^\circ ,180^\circ \}$
D. $\{30^\circ ,270^\circ ,360^\circ \}$
E. $\{30^\circ ,180^\circ ,360^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}-\sin x+\cos x &= 1 \\ \cos x-\sin x &= 1 \end{align}$
$a=1$ (positif) dan $b=-1$ (negatif) maka $\theta $ berada di kuadran IV.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-1}{1} \right) \\ &= {\tan }^{-1}\left( -1 \right) \\ \theta &= 315^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{1^2+(-1)^2} \\ k &= \sqrt{2} \end{align}$
$\begin{align}\cos x-\sin x &= 1 \\ k\cos (x-\theta ) &= c \\ \sqrt{2}\cos (x-315^\circ ) &= 1 \\ \cos (x-315^\circ ) &= \frac{1}{\sqrt{2}} \\ \cos (x-315^\circ ) &= \frac{1}{2}\sqrt{2} \\ \cos (x-315^\circ ) &= \cos 45^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-315^\circ $ dan $g(x)=45^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-315^\circ &= 45^\circ +k.360^\circ \\ x &= 45^\circ +315^\circ +k.360^\circ \\ x &= 360^\circ +k.360^\circ \end{align}$
$k=-1\to x=0^\circ $
$k=0\to x=360^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-315^\circ &= -45^\circ +k.360^\circ \\ x &= -45^\circ +315^\circ +k.360^\circ \\ x &= 270^\circ +k.360^\circ \end{align}$
$k=0\to x=270^\circ $
HP = $\{0^\circ ,270^\circ ,360^\circ \}$
Jawaban: B

Soal No. 7

Himpunan penyelesaian $\sqrt{3}\cos x-\sin x=1$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,300^\circ \}$
B. $\{27^\circ ,270^\circ \}$
C. $\{30^\circ ,270^\circ \}$
D. $\{60^\circ ,270^\circ \}$
E. $\{30^\circ ,60^\circ \}$

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x-\sin x=1$
$a=\sqrt{3}$ (positif) dan $b=-1$ (negatif) maka $\theta $ berada di kuadran IV.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( -\frac{1}{3}\sqrt{3} \right) \\ \theta &= 330^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{3})^2+(-1)^2} \\ k &= 2 \end{align}$
$\begin{align}\sqrt{3}\cos x-\sin x &= 1 \\ k\cos (x-\theta ) &= c \\ 2\cos (x-330^\circ ) &= 1 \\ \cos (x-330^\circ ) &= \frac{1}{2} \\ \cos (x-330^\circ ) &= \cos 60^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-330^\circ $ dan $g(x)=60^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-330^\circ &= 60^\circ +k.360^\circ \\ x &= 390^\circ +k.360^\circ \end{align}$
$k=-1\to x=30^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-330^\circ &= -60^\circ +k.360^\circ \\ x &= 270^\circ +k.360^\circ \end{align}$
$k=0\to x=270^\circ $
HP = $\{30^\circ ,270^\circ \}$
Jawaban: C

Soal No. 8

Himpunan penyelesaian persamaan $\sqrt{3}\cos x-\sin x=\sqrt{2}$ untuk $0^\circ < x < 360^\circ $ adalah ….
A. $\{15^\circ ,75^\circ \}$
B. $\{15^\circ ,285^\circ \}$
C. $\{15^\circ ,315^\circ \}$
D. $\{75^\circ ,285^\circ \}$
E. $\{75^\circ ,315^\circ \}$

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x-\sin x=\sqrt{2}$
$a=\sqrt{3}$ (positif) dan $b=-1$ (negatif) maka $\theta $ berada di kuadran IV.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( -\frac{1}{3}\sqrt{3} \right) \\ \theta &= 330^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{3})^2+(-1)^2} \\ k &= 2 \end{align}$
$\begin{align}\sqrt{3}\cos x-\sin x &= \sqrt{2} \\ k\cos (x-\theta ) &= c \\ 2\cos (x-330^\circ ) &= \sqrt{2} \\ \cos (x-330^\circ ) &= \frac{1}{2}\sqrt{2} \\ \cos (x-330^\circ ) &= \cos 45^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-330^\circ $ dan $g(x)=45^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-330^\circ &= 45^\circ +k.360^\circ \\ x &= 45^\circ +330^\circ +k.360^\circ \\ x &= 375^\circ +k.360^\circ \end{align}$
$k=-1\to x=15^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-330^\circ &= -45^\circ +k.360^\circ \\ x &= -45^\circ +330^\circ +k.360^\circ \\ x &= 285^\circ +k.360^\circ \end{align}$
$k=0\to x=285^\circ $
HP = $\{15^\circ ,285^\circ \}$
Jawaban: B

Soal No. 9

Himpunan penyelesaian persamaan $2\sqrt{3}\sin x-2\cos x=2\sqrt{3}$ pada $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{30^\circ ,150^\circ \}$
B. $\{90^\circ ,150^\circ \}$
C. $\{90^\circ ,180^\circ \}$
D. $\{120^\circ ,150^\circ \}$
E. $\{150^\circ ,180^\circ \}$

Penyelesaian: Lihat/Tutup

$\begin{align}2\sqrt{3}\sin x-2\cos x &= 2\sqrt{3} \\ -2\cos x+2\sqrt{3}\sin x &= 2\sqrt{3} \end{align}$
$a=-2$ (negatif) dan $b=2\sqrt{3}$ (positif) maka $\theta $ di kuadran II.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{2\sqrt{3}}{-2} \right) \\ &= {\tan }^{-1}\left( -\sqrt{3} \right) \\ \theta &= 120^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(-2)^2+(2\sqrt{3})^2} \\ k &= 4 \end{align}$
$\begin{align}-2\cos x+2\sqrt{3}\sin x &= 2\sqrt{3} \\ k\cos (x-\theta ) &= c \\ 4\cos (x-120^\circ ) &= 2\sqrt{3} \\ \cos (x-120^\circ ) &= \frac{2\sqrt{3}}{4} \\ \cos (x-120^\circ ) &= \frac{1}{2}\sqrt{3} \\ \cos (x-120^\circ ) &= \cos 30^\circ \end{align}$
Persamaan trigonometri dasar,$\cos f(x)=\cos g(x)$ dengan $f(x)=x-120^\circ $ dan $g(x)=30^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-120^\circ &= 30^\circ +k.360^\circ \\ x &= 30^\circ +120^\circ +k.360^\circ \\ x &= 150^\circ +k.360^\circ \end{align}$
$k=0\to x=150^\circ $
2) $f(x)=-g(x)+k.360^\circ $
$\begin{align}x-120^\circ &= -30^\circ +k.360^\circ \\ x &= -30^\circ +120^\circ +k.360^\circ \\ x &= 90^\circ +k.360^\circ \end{align}$
$k=0\to x=90$
HP = $\{90^\circ ,150^\circ \}$
Jawaban: B

Soal No. 10

Batas-batas nilai $p$, agar persamaan $(p-2)\cos x+(p-1)\sin x=p$ dapat diselesaikan adalah ….
A. $2\le p\le 3$
B. $1\le p\le 5$
C. $p\le 2$ atau $p\ge 3$
D. $p\le 1$ atau $p\ge 5$
E. $p\le -5$ atau $p\ge 1$

Penyelesaian: Lihat/Tutup

$(p-2)\cos x+(p-1)\sin x=p$
$a=p-2$, $b=p-1$ dan $c=p$
Syarat persamaan dapat diselesaikan adalah:
$\begin{align}a^2+b^2 &\ge c^2 \\ (p-2)^2+(p-1)^2 &\ge p^2 \\ p^2-4p+4+p^2-2p+1 &\ge p^2 \\ p^2-6p+5 &\ge 0 \\ (p-1)(p-5) &\ge 0 \end{align}$
$p\le 1$ atau $p\ge 5$
Jawaban: D

Soal No. 11

Batas-batas nilai $p$ agar persamaan $p\sin x+(p+1)\cos x=p+2$ dapat diselesaikan adalah ….
A. $p\le -1$ atau $p\ge 3$
B. $p\le 1$ atau $p\ge 3$
C. $p\le -3$ atau $p\ge 1$
D. $-1\le p\le 3$
E. $1\le p\le 3$

Penyelesaian: Lihat/Tutup

$p\sin x+(p+1)\cos x=p+2$
$(p+1)\cos x+p\sin x=p+2$
$a=p+1$, $b=p$ dan $c=p+2$
Syarat persamaan dapat diselesaikan adalah:
$\begin{align}a^2+b^2 &\ge c^2 \\ (p+1)^2+p^2 &\ge (p+2)^2 \\ p^2+2p+1+p^2 &\ge p^2+4p+4 \\ p^2-2p-3 &\ge 0 \\ (p+1)(p-3) &\ge 0 \end{align}$
$p\le -1$ atau $p\ge 3$
Jawaban: A

Soal No. 12

Agar persamaan $3\cos x-m\sin x=3\sqrt{5}$ dapat diselesaikan maka nilai $m$ adalah ….
A. $-3\sqrt{6}\le m\le 3\sqrt{6}$
B. $-6\le m\le 6$
C. $0\le m\le 36$
D. $m\le -3\sqrt{6}$ atau $m\ge 3\sqrt{6}$
E. $m\le -6$ atau $m\ge 6$

Penyelesaian: Lihat/Tutup

$3\cos x-m\sin x=3\sqrt{5}$
$a=3$, $b=-m$ dan $c=3\sqrt{5}$
Syarat persamaan dapat diselesaikan adalah:
$\begin{align}a^2+b^2 &\ge c^2 \\ 3^2+(-m)^2 &\ge (3\sqrt{5})^2 \\ 9+m^2 &\ge 45 \\ m^2-36 &\ge 0 \\ (m+6)(m-6) &\ge 0 \end{align}$
Jadi, $m\le -6$ atau $m\ge 6$
Jawaban: E

Soal No. 13

Persamaan $(p-3)\cos x+(p-1)\sin x=p+1$ dapat diselesaikan untuk $p$ dalam batas ….
A. $-9\le p\le -1$
B. $-9\le p\le 1$
C. $1\le p\le 9$
D. $p\le 1$ atau $p\ge 9$
E. $p\le -9$ atau $p\ge 1$

Penyelesaian: Lihat/Tutup

$(p-3)\cos x+(p-1)\sin x=p+1$
$a=p-3$, $b=p-1$ dan $c=p+1$
Syarat persamaan dapat diselesaikan adalah:
$\begin{align}a^2+b^2 &\ge c^2 \\ (p-3)^2+(p-1)^2 &\ge (p+1)^2 \\ p^2-6p+9+p^2-2p+1 &\ge p^2+2p+1 \\ p^2-10p+9 &\ge 0 \\ (p-1)(p-9) &\ge 0 \end{align}$
Jadi, $p\le 1$ atau $p\ge 9$
Jawaban: D

Soal No. 14

Agar persamaan $\sqrt{3}\cos x-\sin x=p$ dapat diselesaikan maka batas-batas nilai $p$ adalah ….
A. $-2\le p\le 2$
B. $-2 < p < 2$
C. $-1\le p\le 1$
D. $-1 < p < 1$
E. $-\sqrt{2}\le p\le \sqrt{2}$

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x-\sin x=p$
$a=\sqrt{3}$, $b=-1$ dan $c=p$
Syarat persamaan dapat diselesaikan adalah:
$\begin{align}a^2+b^2 &\ge c^2 \\ (\sqrt{3})^2+(-1)^2 &\ge p^2 \\ 3+1 &\ge p^2 \\ -p^2+4 &\ge 0 \\ p^2-4 &\le 0 \\ (p+2)(p-2) &\le 0 \end{align}$
Jadi, $-2\le p\le 2$.
Jawaban: B

Soal No. 15

Himpunan penyelesaian persamaan $-3\cos x-\sqrt{3}\sin x=2\sqrt{3}$ untuk $0\le x\le 2\pi $ adalah ….
A. $\left\{ \frac{1}{6}\pi \right\}$
B. $\left\{ \frac{4}{6}\pi \right\}$
C. $\left\{ \frac{5}{6}\pi \right\}$
D. $\left\{ \frac{7}{6}\pi \right\}$
E. $\left\{ \frac{11}{6}\pi \right\}$

Penyelesaian: Lihat/Tutup

$-3\cos x-\sqrt{3}\sin x=2\sqrt{3}$
$a=-3$ (negatif) dan $b=-\sqrt{3}$ (negatif) maka $\theta $ berada di kuadran III.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-\sqrt{3}}{-3} \right) \\ &= {\tan }^{-1}\left( \frac{1}{3}\sqrt{3} \right) \\ \theta &= 210^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{{{(-3)}^2}+{{(-\sqrt{3})}^2}} \\ &= \sqrt{12} \\ k &= 2\sqrt{3} \end{align}$
$\begin{align}-3\cos x-\sqrt{3}\sin x &= 2\sqrt{3} \\ k\cos (x-\theta ) &= c \\ 2\sqrt{3}\cos (x-210^\circ ) &= 2\sqrt{3} \\ \cos (x-210^\circ ) &= 1 \\ \cos (x-210^\circ ) &= \cos 0^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-210^\circ $ dan $g(x)=0^\circ $ maka:
$\begin{align}f(x) &= \pm g(x)+k.360^\circ \\ x-210^\circ &= \pm 0^\circ +k.360^\circ \\ x-210^\circ &= k.360^\circ \\ x &= 210^\circ +k.360^\circ \end{align}$
$k=0\to x=210^\circ =\frac{7}{6}\pi $
Jawaban: D

Soal No. 16

Himpunan penyelesaian persamaan $\sqrt{6}\sin x+\sqrt{2}\cos x=2$ untuk $0^\circ \le x\le 360^\circ $ adalah ….
A. $\{15^\circ ,105^\circ \}$
B. $\{15^\circ ,195^\circ \}$
C. $\{75^\circ ,105^\circ \}$
D. $\{75^\circ ,345^\circ \}$
E. $\{105^\circ ,345^\circ \}$

Penyelesaian: Lihat/Tutup

$\sqrt{6}\sin x+\sqrt{2}\cos x=2$
$\sqrt{2}\cos x+\sqrt{6}\sin x=2$
$a=\sqrt{2}$ (positif) dan $b=\sqrt{6}$ (positif) maka $\theta $ berada di kuadran I.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{\sqrt{6}}{\sqrt{2}} \right) \\ &= {\tan }^{-1}\left( \sqrt{3} \right) \\ \theta &= 60^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{2})^2+(\sqrt{6})^2} \\ &= \sqrt{8} \\ k &= 2\sqrt{2} \end{align}$
$\begin{align}\sqrt{2}\cos x+\sqrt{6}\sin x &= 2 \\ k\cos (x-\theta ) &= c \\ 2\sqrt{2}\cos (x-60^\circ ) &= 2 \\ \cos (x-60^\circ ) &= \frac{1}{\sqrt{2}} \\ \cos (x-60^\circ ) &= \frac{1}{2}\sqrt{2} \\ \cos (x-60^\circ ) &= \cos 45^\circ \end{align}$
Persamaan trigonometri dasar, $\cos f(x)=\cos g(x)$ dengan $f(x)=x-60^\circ $ dan $g(x)=45^\circ $ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-60^\circ &= 45^\circ +k.360^\circ \\ x &= 45^\circ +60^\circ +k.360^\circ \\ x &= 105^\circ +k.360^\circ \end{align}$
$k=0\to x=105^\circ $
2) $f(x)=g(x)+k.360^\circ $
$\begin{align}x-60^\circ &= -45^\circ +k.360^\circ \\ x &= -45^\circ +60^\circ +k.360^\circ \\ x &= 15^\circ +k.360^\circ \end{align}$
$k=0\to x=15^\circ $
HP = $\{15^\circ ,105^\circ \}$
Jawaban: A

Soal No. 17

Bentuk $\cos x+\sin x$ dapat diubah menjadi $k\cos (x-\theta )$. Nilai $k$ dan $\theta $ berturut-turut adalah ….
A. 1 dan $45^\circ $
B. 1 dan $135^\circ $
C. $\sqrt{2}$ dan $45^\circ $
D. $\sqrt{2}$ dan $135^\circ $
E. $\sqrt{2}$ dan $225^\circ $

Penyelesaian: Lihat/Tutup

$\cos x+\sin x$
$a=1$ (positif) dan $b=1$ (positif) maka $\theta $ berada di kuadran I.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{1}{1} \right) \\ &= {\tan }^{-1}\left( 1 \right) \\ \theta &= 45^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{1^2+1^2} \\ k &= \sqrt{2} \end{align}$
Jadi, nilai $k$ dan $\theta $ berturut-turut adalah $\sqrt{2}$ dan $45^\circ $.
Jawaban: C

Soal No. 18

Bentuk $\sqrt{3}\cos x+\sin x$ dapat diubah menjadi bentuk $k\cos (x-A)$ dengan $k>0$ dan $0^\circ \le A\le 360^\circ $ yaitu …..
A. $2\cos (x-30^\circ )$
B. $2\cos (x-60^\circ )$
C. $2\cos (x-45^\circ )$
D. $3\cos (x-30^\circ )$
E. $4\cos (x-30^\circ )$

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x+\sin x$
$a=\sqrt{3}$ (positif) dan $b=1$ (positif) maka $\theta $ di kuadran I.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( \frac{1}{3}\sqrt{3} \right) \\ \theta &= 30^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{3})^2+1^2} \\ k &= 2 \end{align}$
$\begin{align}\sqrt{3}\cos x+\sin x &= k\cos (x-\theta ) \\ &= 2\cos (x-30^\circ ) \end{align}$
Jawaban: A

Soal No. 19

Bentuk $\sqrt{3}\cos x-\sin x$ untuk $0\le x\le \pi $ dapat dinyatakan sebagai ….
A. $2\cos \left( x+\frac{\pi }{6} \right)$
B. $2\cos \left( x+\frac{7\pi }{6} \right)$
C. $2\cos \left( x+\frac{11\pi }{6} \right)$
D. $2\cos \left( x-\frac{7\pi }{6} \right)$
E. $2\cos \left( x-\frac{\pi }{6} \right)$

Penyelesaian: Lihat/Tutup

$\sqrt{3}\cos x-\sin x$
$a=\sqrt{3}$ (positif) dan $b=-1$ (negatif) maka $\theta $ di kuadran IV.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-1}{\sqrt{3}} \right) \\ &= {\tan }^{-1}\left( -\frac{1}{3}\sqrt{3} \right) \\ \theta &= 330^\circ \\ \theta &= \frac{11\pi }{6} \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{(\sqrt{3})^2+(-1)^2} \\ k &= 2 \end{align}$
$\begin{align}\sqrt{3}\cos x-\sin x &= k\cos (x-\theta ) \\ &= 2\cos \left( x-\frac{11\pi }{6} \right) \end{align}$
Ubah menggunakan Perbandingan Trigonometri Sudut Berelasi, $\cos A=\cos (2\pi +A)$ maka $\begin{align}\sqrt{3}\cos x-\sin x &= 2\cos \left( x-\frac{11\pi }{6} \right) \\ &= 2\cos \left( 2\pi +x-\frac{11\pi }{6} \right) \\ \sqrt{3}\cos x-\sin x &= 2\cos \left( x+\frac{\pi }{6} \right) \end{align}$
Jawaban: A

Soal No. 20

Bentuk $-3\cos x-\sqrt{3}\sin x$ dapat dinyatakan dalam ….
A. $2\sqrt{3}\cos (x-150^\circ )$
B. $2\sqrt{3}\cos (x-210^\circ )$
C. $-2\sqrt{3}\cos (x-210^\circ )$
D. $-2\sqrt{3}\cos (x-30^\circ )$
E. $2\sqrt{3}\cos (x-30^\circ )$

Penyelesaian: Lihat/Tutup

$-3\cos x-\sqrt{3}\sin x$
$a=-3$ (negatif) dan $b=-\sqrt{3}$ (negatif) maka $\theta $ berada di kuadran III.
$\begin{align}\theta &= {\tan }^{-1}\left( \frac{b}{a} \right) \\ &= {\tan }^{-1}\left( \frac{-\sqrt{3}}{-3} \right) \\ &= {\tan }^{-1}\left( \frac{1}{3}\sqrt{3} \right) \\ \theta &= 210^\circ \end{align}$
$\begin{align}k &= \sqrt{a^2+b^2} \\ &= \sqrt{{{(-3)}^2}+(-\sqrt{3})^2} \\ &= \sqrt{12} \\ k &= 2\sqrt{3} \end{align}$
$\begin{align}-3\cos x-\sqrt{3}\sin x &= k\cos (x-\theta ) \\ &= 2\sqrt{3}\cos (x-210^\circ ) \end{align}$
Jawaban: B

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