Trigonometrical Equations 1 Question 16
16. The general solution of $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$ is
(1989, 2M)
(a) $n \pi+\frac{\pi}{8}$
(b) $\frac{n \pi}{2}+\frac{\pi}{8}$
(c) $(-1)^{n} \frac{n \pi}{2}+\frac{\pi}{8}$
(d) $2 n \pi+\cos ^{-1} \frac{3}{2}$
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Answer:
Correct Answer: 16. (b)
Solution:
- Given, $\sin 3 x+\sin x-3 \sin 2 x=\cos 3 x+\cos x-3 \cos 2 x$
$\Rightarrow 2 \sin 2 x \cos x-3 \sin 2 x=2 \cos 2 x \cos x-3 \cos 2 x$
$\Rightarrow \quad \sin 2 x(2 \cos x-3)=\cos 2 x(2 \cos x-3)$
$\Rightarrow \sin 2 x =\cos 2 x $
$\Rightarrow \tan 2 x =1 $
$\Rightarrow 2 x =n \pi+\frac{\pi}{4} \quad \Rightarrow \quad x=\frac{n \pi}{2}+\frac{\pi}{8}$
