The general solution of
$sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x$ is

n π + \frac{π}{8}

where

n ϵ I

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\frac{n π}{2} + \frac{π}{8}

where

n ϵ I

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(- 1)^{n} (\begin{matrix} \frac{n π}{2} + \frac{π}{8} \end{matrix})

where

n ϵ I

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2 n π + c o s^{- 1} (\begin{matrix} \frac{3}{2} \end{matrix})

where

n ϵ I

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Solution

The correct option is B $\frac{n π}{2} + \frac{π}{8}$ where $n ϵ I$
Given equation is
$sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x$
$(sin x + sin 3 x) - 3 sin 2 x = (cos x + cos 3 x) - 3 cos 2 x$
$(2 sin 2 x cos x - 3 sin 2 x) - (2 cos 2 x cos x - 3 cos 2 x) = 0$
$sin 2 x (2 cos x - 3) - cos 2 x (2 cos x - 3) = 0$
$\Rightarrow (sin 2 x - cos 2 x) (2 cos x - 3) = 0$
$c o s x = \frac{3}{2}$
which is not possible
$sin 2 x = cos 2 x$
$tan 2 x = 1$
$tan 2 x = tan \frac{π}{4}$
$2 x = n π + \frac{π}{4}$
$\Rightarrow x = \frac{n π}{2} + \frac{π}{8}, n \in I$

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