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

n π + \frac{π}{8}

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\frac{n π}{2} + \frac{π}{8}

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2 n π + {cos}^{- 1} (\frac{1}{3})

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2 n π + {cos}^{- 1} (\frac{3}{2})

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Solution

The correct option is B $\frac{n π}{2} + \frac{π}{8}$
$s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x$
$\Rightarrow s i n x - 6 s i n x c o s x + 3 s i n x - 4 s i n^{2} x = c o s x$
Given
$s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x$
$\Rightarrow s i n 3 x + s i n x - 3 s i n 2 x = c o s 3 x + c o s x - 3 c o s 2 x$
$\Rightarrow 2 s i n 2 x c o s x - 3 s i n 2 x = 2 c o s 2 x c o s x - 3 c o s 2 x$
$\Rightarrow 2 s i n 2 x c o s x - 2 c o s 2 x c o s x = 3 s i n 2 x - 3 c o s 2 x$
$\Rightarrow 2 c o s x (s i n 2 x - c o s 2 x) = 3 (s i n 2 x - c o s 2 x)$
$\Rightarrow (2 c o s x - 3) (s i n 2 x - c o s 2 x) = 0$
$∴ c o s x = \frac{3}{2},$ or $s i n x = c o s 2 x$
$↓$
Impossible $\Rightarrow t a n 2 x = 1$
$\Rightarrow t a n 2 x = t a n \frac{π}{4}$
$\Rightarrow 2 x = x π + \frac{π}{4}$
$\Rightarrow x = \frac{x π}{2} + \frac{π}{8}$
Hence, Genral solution is $x = \frac{x π}{2} + \frac{π}{8}$

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