$2\sin^2x-3\cos^2x+1=0$
$\Leftrightarrow 2-2\cos^2x-3\cos^2x+1=0$
$\Leftrightarrow \cos^2x=\dfrac{3}{5}$
$\Leftrightarrow \dfrac{1+\cos2x}{2}=\dfrac{3}{5}$
$\Leftrightarrow \cos2x=\dfrac{1}{5}$
$\Leftrightarrow x=\pm\dfrac{1}{2}\arccos\dfrac{1}{5}+k\pi$