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
 A: 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?
A: 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 $x = 1.15$
A: This is only a partial solution, showing there are definitely two solutions and probably no others. It also indicates where the computational difficulty lies in providing a complete answer.
The equation can be rewritten as
$$1+2\sin x\cos x=\sin x+1-\cos^2x$$
which simplifies to
$$\cos^2x=\sin x(1-2\cos x)$$
or, on dividing each side by both $\cos x$ and $\sin x$, 
$$\cot x=\sec x-2$$
If you now draw even a crude sketch of the cotangent and secant function (shifted down by $2$), you can see they intersect once in the interval $(0,\pi/2)$, once in the interval $(\pi/2,\pi)$, and not at all in the interval $(\pi,3\pi/2)$. (That is, $\cot x$ decreases from $\infty$ to $0$ while $\sec x-2$ increases from $-1$ to $\infty$ in $(0,\pi/2)$, etc.)
The final interval, $(3\pi/2,2\pi)$ is trickier, because both curves are decreasing there. Using special values of trig functions at multiples of $\pi/4$ and $\pi/6$, you can limit any possible intersections to the interval $(5\pi/3,7\pi/4)$, since $\cot x$ decreases from $-1/\sqrt3\approx-.577$ to $-1$ on that interval while $\sec x-2$ decreases from $0$ to $1/\sqrt2-1\approx-.586$. The problem is, $-.586\lt-.577$, so we can't definitely say the curves don't intersect. Presumably computing at the midpoint, $x=41\pi/24$, would be enough to show the curves don't cross, but that seems like a lot of work; there ought to be a better way. (Drawing the graphs of $y=\cot x$ and $y=\sec x-2$ carefully will, of course, settle the matter, but that's a lot of work too! If you rely on a computer to do the work, you might as well just ask the computer to graph the curve $y=1+\sin2x-\sin x-\sin^2x$.)
A: Use the tan-half-angle substitution ($t \rightarrow \tan\left( \frac{x}{2} \right)$) to turn the equation into a polynomial.
$$ \begin{cases} 
  \sin(x)  = \frac{2 t}{1+t^2} \\
  \cos(x) = \frac{1-t^2}{1+t^2} \\
  x  =2 \tan^{-1}(t)
\end{cases} $$
Since $\sin(2 x) = 2 \sin(x) \cos(x)$ the expression is 
$$ 1 + 2 \frac{2 t}{1+t^2} \frac{1-t^2}{1+t^2} = \frac{2 t}{1+t^2} + \left( \frac{2 t}{1+t^2} \right)^2 $$
or 
$$ \left. \frac{2 t (3 t^2+2 t-1)}{(1+t^2)^2} = 1 \right\} (1+t^2)^2 -2 t ( 3 t^2+2 t-1) =0$$
This polynomial is of 4th order with two real solutions

$$ \begin{cases} t = 0.647989325215799 & \rightarrow x = 2.82499129600076
 \\
t = 6.26423704911644 & \rightarrow x = 1.14992088410486
\end{cases} $$

