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

Do you remember the double angle formulas? I don't see you using that.
Now, consider writing it all again using this fact. I've attached my work that get you to a point where you need to reason out the angles, and I've hidden it so you can at least work out some more using the double angle formula! :)

 $$ 1 + \sin(2x) = \sin(x) + \sin^2(x)\\1 + 2\sin(x)\cos(x) = \sin^2(x) + \sin(x)\\ 2\sin(x)\cos(x) = \sin^2(x) - 1 + \sin(x)\\ 2\sin(x)\cos(x) = -\cos^2(x) + \sin(x)\\ 2\sin(x)\cos(x) + \cos^2(x) = \sin(x)\\ 2\sin(x) + \cos(x) = \tan(x)\\ 2 + \cot(x) = \tan(x)\csc(x)\\ 2 + \cot(x) = \sec(x)\\ 2 = \sec(x) - \cot(x)\\ 2 = \frac{1}{\cos(x)} - \frac{\cos(x)}{\sin(x)} $$

And from there you should be able to reason out the angles which satisfy that equality.
Edit: Maybe you wrote the problem down wrong for us?
