Question #da791

1 Answer
Aug 22, 2016

#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.

wikipedia.org

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#