If 0

Question

(C) Given equation: $\sin 	heta - 3 \sin 2 	heta + \sin 3 	heta = \cos 	heta - 3 \cos 2 	heta + \cos 3 	heta$ Rearranging the terms: $(\sin 	he

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$\frac{\pi}{8}$

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(C) Given equation: $\sin 	heta - 3 \sin 2 	heta + \sin 3 	heta = \cos 	heta - 3 \cos 2 	heta + \cos 3 	heta$ Rearranging the terms: $(\sin 	heta + \sin 3 	heta) - 3 \sin 2 	heta = (\cos 	heta + \cos 3 	heta) - 3 \cos 2 	heta$ Using sum-to-product formulas $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$: $2 \sin 2 	heta \cos 	heta - 3 \sin 2 	heta = 2 \cos 2 	heta \cos 	heta - 3 \cos 2 	heta$ $\sin 2 	heta (2 \cos 	heta - 3) = \cos 2 	heta (2 \cos 	heta - 3)$ $(\sin 2 	heta - \cos 2 	heta)(2 \cos 	heta - 3) = 0$ Since $2 \cos 	heta - 3 = 0 \Rightarrow \cos 	heta = \frac{3}{2}$,which is impossible as $-1 \leq \cos 	heta \leq 1$. Therefore,$\sin 2 	heta - \cos 2 	heta = 0$ $\sin 2 	heta = \cos 2 	heta \Rightarrow 	an 2 	heta = 1$ Since $0 $	an 2 	heta = 	an \frac{\pi}{4}$ $\Rightarrow 2 	heta = \frac{\pi}{4}$ $\Rightarrow 	heta = \frac{\pi}{8}$.

If $0 < \theta < \frac{\pi}{2}$,then the solution of the equation $\sin \theta - 3 \sin 2 \theta + \sin 3 \theta = \cos \theta - 3 \cos 2 \theta + \cos 3 \theta$ is

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