+130561 cos^3 (2x) = cos(2x) subtract cos(2x) from each side
cos^3 (2x) - cos(2x) = 0 factor out cos(2x)
cos(2x) [ cos^2 (2x) - 1 ] = 0
Setting both factors to 0 and solving, we have
cos(2x) = 0
cos(x) = 0 at x = pi/2 , 3pi/2, 5pi/2 and at 7pi/2
Then cos(2x) will = 0 when x = pi/4 , 3pi/4, 5pi/4, 7pi/4
For the second factor, we have
cos^(2x) - 1 = 0 factor as a difference of squares
[cos (2x) - 1] [ cos (2x) + 1] =0
Setting the first factor to 0, we have
cos(2x) - 1 = 0 add 1 to both sides
cos(2x) = 1
cos(x) = 1 when x = 0 and when x = 2pi
So .... cos(2x) will = 1 when x = 0 and x = pi
And for the last factor, we have
cos(2x) + 1 = 0 subtract 1 from both sides
cos(2x) = -1
cos(x) = -1 when x = pi and when x = 3pi
So.......cos(2x) will = 0 when x = pi/2 and x = 3pi/2
So....the solutions are
x =0 , x = pi/4 , x = pi/2 , x = 3pi/4 ,x = pi, x= 5pi/4, x = 3pi/2, x = 7pi/4
Here's a graph :
https://www.desmos.com/calculator/5zuwwk5xwb
