Answer
$(\frac{\pi}{2}+2n{\pi},3)$ and $(\frac{3\pi}{2}+2n{\pi},-1)$ where n is any integer
Work Step by Step
For the tangent line to be horizontal
$f'(x)$ = $0$
$f(x)$ = $2sinx+sin^{2}x$
$f'(x)$ = $2cosx+2sinxcosx$
$0$ = $2cosx+2sinxcosx$
$0$ = $2cosx(1+2sinx)$
$cosx$ = $0$
$x$ = $\frac{\pi}{2}+2n{\pi}$
$1+2sinx$ = $0$
$sinx$ = $-\frac{1}{2}$
$x$ = $\frac{3\pi}{2}+2n{\pi}$
$f(\frac{\pi}{2})$ = $3$
$f(\frac{3\pi}{2})$ = $-1$
so the point on the curve with a horizontal tangent are $(\frac{\pi}{2}+2n{\pi},3)$ and $(\frac{3\pi}{2}+2n{\pi},-1)$ where n is any integer
