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))

 

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