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.