Question

Question #da791

Answer 1

`x = pi/4+npi/2, n in ZZ`

Explanation:

`2sin^2(x) = 1`

`=> sin^2(x) = 1/2`

`=> sin(x) = +-sqrt(1/2) = +-1/sqrt(2) = +-sqrt(2)/2`

If we look at a unit circle, we will find that `|sin(x)|=sqrt(2)/2` at an angle of `pi/4` in each quadrant.

As we want both positive and negative values for `sin(x)`, we find that `pi/4` plus any integer multiple of `pi/2` is a solution. Thus we get the solution set

`x = pi/4+npi/2, n in ZZ`