Question

How do you solve `2sin^2x=1`?

Answer 1

`S={x|x=pi/4+k/2pi " where " kinRR}`.

Explanation:

`2sin^2x=1`
`sin^2x=1/2`
`sinx=+-sqrt(1/2)`
`sinx=+-sqrt(2)/2`
This is a common value that we get with `x=pi/4,3/4pi,5/4pi,7/4pi,...`
So, the solutions is:
`S={x|x=pi/4+k/2pi " where " kinRR}`.