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.

 

[Architeuthis Dux]

for your question. The notation sin^2 x is another way of writing (sin x)^2, which means the sine of x squared. So when the problem says sin^2 x = 1/4, it is asking you to find all values of x where the sine of x squared is equal to 1/4. This can be solved by taking the square root of both sides, which gives you sin x = 1/2. From there, you can use the unit circle or trigonometric identities to find the solutions in the given range of pi/2 <= x < 2pi. There is no need to use the Pythagorean theorem in this case. I hope this helps clarify the problem for you.