$\sin^2x+\sin2x-2\cos^2x=\dfrac{1}{2}$
$⇔\sin^2x+\sin2x-2\cos^2x=\dfrac{1}{2}.1$
$⇔\sin^2x+2\sin x\cos x-2\cos^2x=\dfrac{1}{2}.(\sin^2x+\cos^2x)$
$⇔\sin^2x+2\sin x\cos x-2\cos^2x=\dfrac{1}{2}\sin^2x+\dfrac{1}{2}\cos^2x$
$⇔\left(\sin^2x-\dfrac{1}{2}\sin^2x\right)+2\sin x\cos x-\left(2\cos^2x+\dfrac{1}{2}\cos^2x\right)=0$
$⇔\dfrac{1}{2}\sin^2x+2\sin x\cos x-\dfrac{5}{2}\cos^2x=0$
$⇔\sin^2x+4\sin x\cos x-5\cos^2x=0$.
