You can solve for [tex]\sin^2\theta[/tex], then take square roots:
[tex]2\sin^2\theta=1\implies\sin^2\theta=\dfrac12\implies\sin\theta=\pm\dfrac1{\sqrt2}[/tex]
Then
[tex]\theta=\dfrac\pi4+2n\pi,\dfrac{3\pi}4+2n\pi,-\dfrac\pi4+2n\pi,\text{ or }-\dfrac{3\pi}4+2n\pi[/tex]
where [tex]n[/tex] is any integer; we get the following solutions in the interval [tex]0\le\theta\le2\pi[/tex]:
[tex]\boxed{\theta=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4}[/tex]
Or, you can use the double angle identity:
[tex]\sin^2\theta=\dfrac{1-\cos(2\theta)}2=\dfrac12\implies1-\cos(2\theta)=1\implies\cos(2\theta)=0[/tex]
Then
[tex]2\theta=\dfrac\pi2+n\pi\implies\theta=\dfrac\pi4+\dfrac{n\pi}2[/tex]
where [tex]n[/tex] is any integer, and we get the same solutions as above within the prescribed interval.
