$2\cos^2 2x+3\sin^2x=2\\⇔2(1-2\sin^2x)^2+3\sin^2x=2\\⇔2(4\sin^4x-4\sin^2x+1)+3\sin^2x-2=0\\⇔8\sin^4x-5\sin^2x=0\\⇔$\(\left[ \begin{array}{l}\sin{x}=0\\\sin{x}=±\sqrt{\dfrac{5}{8}}\end{array} \right.\) $\\$$⇔$\(\left[ \begin{array}{l}x=kπ\\x=\arcsin(±\sqrt{\dfrac{5}{8}})+k2π\\x=π-\arcsin(±\sqrt{\dfrac{5}{8}})+k2π\end{array} \right.\) $(k\in\mathbb{Z})$
