Mathematics - Trigonometric Equation Question with Solution | TestHub

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