Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ in $x\in(0,2\pi)$
solution i try $1+\sin 2x =\sin x+\sin^2 x$
$5+4\sin 2x=4\sin^2 x+4\sin x+1$
$5+\sin 2x=(2\sin x+1)^2$ How i solve it after that point

Assume $y = \sin x$ 
$$1 + 2y\sqrt{1-y^2} = y + y^2 $$
$$2y\sqrt{1-y^2} = (y + y^2 -1) \implies 4y^2(1-y^2) = (y+ y^2 -1)^2$$
Simplifying, we get $5y^4 + 2y^3 - 5y^2 -2y + 1 = 0$
Solve somehow to get $y = 0.311$ or $y = 0.913$ and so $x = 0.316$ or $x = 1.15$
