If sin 5x + sin 3x + sin x = 0,then the value | vedclass

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(C) Given equation: $\sin 5x + \sin 3x + \sin x = 0$Grouping the terms: $(\sin 5x + \sin x) + \sin 3x = 0$Using the formula $\sin C + \sin D = 2 \sin

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(C) Given equation: $\sin 5x + \sin 3x + \sin x = 0$Grouping the terms: $(\sin 5x + \sin x) + \sin 3x = 0$Using the formula $\sin C + \sin D = 2 \sin \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$:$2 \sin 3x \cos 2x + \sin 3x = 0$$\sin 3x (2 \cos 2x + 1) = 0$This implies either $\sin 3x = 0$ or $2 \cos 2x + 1 = 0$.Case $1$: $\sin 3x = 0$ $\Rightarrow 3x = n\pi$ $\Rightarrow x = \frac{n\pi}{3}$. For $0 Case $2$: $2 \cos 2x = -1 \Rightarrow \cos 2x = -\frac{1}{2}$.Since $\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}$,we have $2x = 2n\pi \pm \frac{2\pi}{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}$.For $0 Thus,the value of $x$ is $\frac{\pi}{3}$.

If $\sin 5x + \sin 3x + \sin x = 0$,then the value of $x$ other than $0$ lying between $0 \le x \le \frac{\pi}{2}$ is

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