The number of solutions of 3x=2x in the interval (2,) is:

Question

The number of solutions of $\sin3x=\cos2x$ in the interval $\left(\dfrac\pi2,\pi\right)$ is:

Accepted Answer

$$3\sin x-4\sin^3x = 1-2\sin^2x \implies 4\sin^3x-2\sin^2x-3\sin x+1=0$$

This factors to give $\sin x = 1$ or $\sin x = \dfrac{-2\pm2\sqrt5}8 = \dfrac{-1\pm\sqrt5}4$.

In $\left(\tfrac\pi2,\pi\right)$: only $\sin x = \dfrac{-1+\sqrt5}4$ gives a valid solution --- 1 solution.

Solutions · 1