Answer:
[tex]x=0,x=\pi,x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}[/tex]
Step-by-step explanation:
This is a trigonometric equation where we need to use the cosine of the double-angle formula
[tex]cos4x=2cos^22x-1[/tex]
Replacing in the equation
[tex]cos4x - cos2x = 0[/tex]
We have
[tex]2cos^22x-1 - cos 2x = 0[/tex]
Rearranging
[tex]2cos^22x - cos 2x-1 = 0[/tex]
A second-degree equation in cos2x. The solutions are:
[tex]cos2x=1,cos2x=-\frac{1}{2}[/tex]
For the first solution
cos2x=1 we find two solutions (so x belongs to [0,2\pi))
[tex]2x=0, 2x=2\pi[/tex]
Which give us
[tex]x=0,x=\pi[/tex]
For the second solution
[tex]cos2x=-\frac{1}{2}[/tex]
We find four more solutions
[tex]2x=\frac{2\pi}{3},2x=\frac{4\pi}{3},2x=\frac{8\pi}{3},2x=\frac{10\pi}{3}[/tex]
Which give us
[tex]x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}[/tex]
All the solutions lie in the interval [tex][0,2\pi)[/tex]
Summarizing. The six solutions are
[tex]x=0,x=\pi,x=\frac{\pi}{3},x=\frac{2\pi}{3},x=\frac{4\pi}{3},x=\frac{5\pi}{3}[/tex]
