Start with
[tex]\dfrac{1}{\sin^2(x)}=4[/tex]
Invert both sides:
[tex]\sin^2(x) = \dfrac{1}{4}[/tex]
Consider the square root of both sides with double sign:
[tex]\sin(x) = \pm\sqrt{\dfrac{1}{4}} = \pm\dfrac{1}{2}[/tex]
We have
[tex]\sin(x) = \dfrac{1}{2} \iff x = \dfrac{\pi}{6}\ \lor\ x = \dfrac{5\pi}{6}[/tex]
and
[tex]\sin(x) = -\dfrac{1}{2} \iff x = \dfrac{7\pi}{6}\ \lor\ x = \dfrac{11\pi}{6}[/tex]
