$P=2-2\sin x-\cos2x$
$=2-2\sin x-(1-2\sin^2x)$
$=2-2\sin x-1+2\sin^2x$
$=2\sin^2x-2\sin x+1$
$=2\Big(\sin^2x-\sin x+\dfrac{1}{2}\Big)$
$=2\Big(\sin x-\dfrac{1}{2}\Big)^2+\dfrac{1}{2}\ge \dfrac{1}{2}$
Vậy $P\min=\dfrac{1}{2}\Leftrightarrow \sin x=\dfrac{1}{2}$