The number of solutions of $\sin 3 x=\cos 2 x$, in the interval $\left(\frac{\pi}{2}, \pi\right)$ is

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Aicharya · Accepted Answer

$\sin 3 x=\cos 2 x$ $ \begin{aligned} &\Rightarrow 3 \sin x-4 \sin ^3 x=1-2 \sin ^2 x \ &\Rightarrow 4 \sin ^3 x-2 \sin ^2 x-3 \sin x+1=0 \ &\Rightarrow \sin x=1, \frac{-2 \pm 2 \sqrt{5}}{8} \end{aligned} $ In the interval $\left(\frac{\pi}{2}, \piight), \sin x=\frac{-2+2 \sqrt{5}}{8}$ So, there is only one solution

The number of solutions of $\sin 3 x=\cos 2 x$, in the interval $\left(\frac{\pi}{2}, \pi\right)$ is

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