The solution of the equation $y\sin x\frac{dy}{dx}=\cos x(\sin x-\frac{y^2}{2})$, given y = 1 when $x=\frac{\pi}{2}$ is:

$y^2=\sin x$

$y\sin x\frac{dy}{dx}=\cos x(\sin x-\frac{y^2}{2})$

$⇒y\frac{dy}{dx}+\frac{y^2\cos x}{2\sin x}=\cos x$

Substitute $y^2=t⇒2y\frac{dy}{dx}=\frac{dt}{dx}⇒\frac{dt}{dx}+t\cot x=2\cos x$  [Linear D.E.]

$I.F.=e^{\int\cot xdx}=\sin x;t.(\sin x)=2\int\cos x\sin x dx +c$

$⇒y^2(\sin x)=\sin ^2x+c$

Since curve passes through $(\frac{\pi}{2},1)⇒c=0$

$⇒y^2=\sin x$