Solve the equation
cos2x−cosx=0
on the interval [0,2π)

cos 2x – cosx = 0 (1)

where cos2x = 2cos^2x – 1

equation (1) becomes;

2cos^2x – 1 – cosx  = 0 (2)

Suppose cosx = y, then equation (2) becomes;

2y^2 – 1 – y = 0

By factorization method;

(2y + 1) x (y – 1) = 0

Where 2y + 1 = 0 or y – 1 = 0

So solving both;

y = -1/2 or 1

As y = cosx

So cosx = -1/2 or 1

The above equation is true only for x = 2/3π, 0, 4/3 π

The solution were find out for interval [0,2 π), where 2 π is not included in the interval, as the interval is not a closed on the right side.

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