# solving for x

• Nov 7th 2009, 03:46 PM
samtheman17
solving for x
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.....
• Nov 7th 2009, 03:58 PM
VonNemo19
Quote:

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

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

If so

Solve the equation, and then restrict the infinite amount of solutions to the given interval. It's lookin like $\displaystyle \sin{x}=\pm\frac{\sqrt{2}}{2}$
• Nov 7th 2009, 04:02 PM
samtheman17
okay thanks,

but how do you solve this equation?
• Nov 7th 2009, 04:13 PM
VonNemo19
Quote:

Originally Posted by samtheman17
okay thanks,

but how do you solve this equation?

$\displaystyle 2\sin^2x=1$

$\displaystyle \sin^2x=\frac{1}{2}$

$\displaystyle \sin{x}=\pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2 }$

$\displaystyle x=\arcsin{\pm\frac{\sqrt{2}}{2}}$