Let [tex]t=\sin(x)[/tex]. The equation becomes
[tex]t^2-2t=0 \iff t(t-2)=0[/tex]
So, the solutions are
[tex]t=0,\quad t-2=0 \iff t=2[/tex]
Substitute back [tex]t=\sin(x)[/tex], we have
[tex]\sin(x)=0,\quad \sin(x)=2[/tex]
Since the sine function is bounded between -1 and 1, it can never equal 2. So, the only possible solution is
[tex]\sin(x)=0 \iff x = k\pi,\quad k \in \mathbb{Z}[/tex]
Depending on the interval you're interested in (you didn't write it!) select the solution(s) that applies(apply)
