Solving sin²x = 1/4 in Quadrants 2, 3, and 4

[alpha01]

find all solutions to sin^2 x = 1/4 in pi/2 <= x < 2pi (i.e in quadrants 2,3 and 4)


i understand what i need to do but don't understand sin^2 part. will i need to apply pythagoras therom here first?



thanks

 

[Snazzy]

$$\sin ^2x=(\sin x)^2$$

You can square root both sides, and finding x should be simple then.