solving for x

**samtheman17** · November 7th 2009, 03:46 PM

solve the following for x:

2sin^2=1 (0 < x < 2*%pi)

what is the solution set??

i think there are 4 answers to this , but im not sure how to do this.....

**VonNemo19** · November 7th 2009, 03:58 PM

Originally Posted by samtheman17

solve the following for x:

2sin^2=1 (0 < x < 2*%pi)

what is the solution set??

i think there are 4 answers to this , but im not sure how to do this.....

Do you mean

$2\sin^2{x}=1\text{ for }(0<x<2\pi)$ ?

If so

Solve the equation, and then restrict the infinite amount of solutions to the given interval. It's lookin like $\sin{x}=\pm\frac{\sqrt{2}}{2}$