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

n π + \frac{π}{8}

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\frac{n π}{2} + \frac{π}{8}

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(- 1)^{n} \frac{n π}{2} + \frac{π}{8}

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2 n π + c o s^{- 1} (\frac{3}{4})

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Solution

The correct option is B $\frac{n π}{2} + \frac{π}{8}$

2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x
$\Rightarrow s i n 2 x = c o s 2 x \Rightarrow T a n 2 x = 1$

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Similar questions

sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x

sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x

n π / 2 + π / 8

sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x

0 \leq x \leq 2 π,

x =

sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x

0 \leq x \leq 2 π

sin x + 3 sin 2 x + sin 3 x = cos x + 3 cos 2 x + cos 3 x

0 \leq x \leq 2 π

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