SATHEE: Trigonometrical Equations 1 Question 16

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) $n π + \frac{π}{8}$

(b) $\frac{n π}{2} + \frac{π}{8}$

(c) $(- 1)^{n} \frac{n π}{2} + \frac{π}{8}$

(d) $2 n π + \cos^{- 1} \frac{3}{2}$

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Answer:

Correct Answer: 16. (b)

Solution:

Given, $\sin 3 x + \sin x - 3 \sin 2 x = \cos 3 x + \cos x - 3 \cos 2 x$

$\Rightarrow 2 \sin 2 x \cos x - 3 \sin 2 x = 2 \cos 2 x \cos x - 3 \cos 2 x$

$\Rightarrow \sin 2 x (2 \cos x - 3) = \cos 2 x (2 \cos x - 3)$

$\Rightarrow \sin 2 x = \cos 2 x$

$\Rightarrow \tan 2 x = 1$

$\Rightarrow 2 x = n π + \frac{π}{4} \Rightarrow x = \frac{n π}{2} + \frac{π}{8}$

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