Question

Not a hard question i just don't get it. I really need some help here, its the final

Answer 1

Start with

$\dfrac{1}{\sin^2(x)}=4$

Invert both sides:

$\sin^2(x) = \dfrac{1}{4}$

Consider the square root of both sides with double sign:

$\sin(x) = \pm\sqrt{\dfrac{1}{4}} = \pm\dfrac{1}{2}$

We have

$\sin(x) = \dfrac{1}{2} \iff x = \dfrac{\pi}{6}\ \lor\ x = \dfrac{5\pi}{6}$

and

$\sin(x) = -\dfrac{1}{2} \iff x = \dfrac{7\pi}{6}\ \lor\ x = \dfrac{11\pi}{6}$