<s>[INST] 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
 [/INST] 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

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

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