The general solution of $sin x - 3 sin 2 x + sin 3 x = cos x - 3 cos 2 x + cos 3 x$ in the interval $0 \leq x \leq 2 π,$ then find $x =$

\frac{π}{8}

\frac{5 π}{8}

\frac{2 π}{3}

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\frac{π}{8}

\frac{5 π}{8}

\frac{9 π}{8}

\frac{13 π}{8}

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\frac{4 π}{3}

\frac{9 π}{3}

\frac{2 π}{3}

\frac{13 π}{8}

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\frac{π}{8}

\frac{5 π}{8}

\frac{9 π}{3}

\frac{4 π}{3}

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Solution

The correct option is B $\frac{π}{8}$ , $\frac{5 π}{8}$ , $\frac{9 π}{8}$ , $\frac{13 π}{8}$
$(s i n x - s i n 3 x) - (c o s x - c o s 3 x) + 3 (s i n 2 x - c o s 2 x) = 0$

$\Rightarrow 2 s i n 2 x . c o s x - 2 c o s 2 x . c o s x + 3 (s i n 2 x - c o s 2 x) = 0$

$\Rightarrow 2 c o s x (s i n 2 x - c o s 2 x) + 3 (s i n 2 x - c o s 2 x) = 0$

$\Rightarrow (2 c o s x + 3) (s i n 2 x - c o s 2 x) = 0$

$c o s x \neq - \frac{3}{2}$

$t h u s sin 2 x = cos 2 x$

$o r, tan 2 x = 1$

$2 x = n π + \frac{π}{4}$

$x = \frac{n π}{2} + \frac{π}{8}$

$h e n c e x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}, \frac{13 π}{8}$

Suggest Corrections

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s i n^{4} \frac{π}{8} + s i n^{4} \frac{3 π}{8} + s i n^{4} \frac{5 π}{8} + s i n^{4} \frac{7 π}{8} =

{cos}^{4} \frac{π}{8} + {cos}^{4} \frac{3 π}{8} + {cos}^{4} \frac{5 π}{8} + {cos}^{4} \frac{7 π}{8}

{cos}^{4} \frac{π}{8} + {cos}^{4} \frac{3 π}{8} + {cos}^{4} \frac{5 π}{8} + {cos}^{4} \frac{7 π}{8}

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