Hi, Find all the solutions between 0 and 2pi to the equation cos(2x)=sin(x)+1 = 1 - 2sin^2 1 - 2sin^2(x) = sin (x) + 1 2sin^2(x) + sin (x) =0 sin(x)[2sin(x) + 1] = 0 sin(x)=0 x = 0, 2sin(x) + 1 = 0 sin(x) = -1/2 x = 7/6 pi, 11/6 pi (cos(x), sin(x))
