recall the identity cos2α = 1 - 2(sinα)^2
so cosx = 1 - 2(sin(x/2))^2
so , sin(x/2) + 1 - 2(sin(x/2))^2 - 1 = 0 now we have a quadratic to solve
so [sin(x/2)][ 1 - 2(sin(x/2) ] = 0
so sin(x/2) = 0 or sin(x/2) = 1/2
so x = 0 , π , 2π , 3π or x = π / 12 , 5π / 12 , 13π / 12 , 17π / 12
hope this helps