cos^6x = cos^4x subtract cos^4x ffrom both sides
cos^6x - cos^4x = 0 faactor this
cos^4x*(cos^2x - 1) = 0
cos^4x (cosx + 1)(cosx-1) setting each factor to 0, we have
x = 0, pi/2, pi , (3 pi)/2 in the interval [0, 2pi)
More general solutions are x = 0 + (pi)n and x = pi/2 + (pi/2)n where n is an integer
$$\\cos^6x=cos^4x\\
$Divide both sides by $\;cos^4x\\
cos^2x=1\\
cosx=\pm1
$At this point I just think about the unit circle$\\
$cos of an angle is given by the x value (not the angle x)$\\
$The x value is 1 at 0 and it is -1 at 180 degrees$\\
So \\
x=180n$ degrees where $ n\in Z\\
or\\
x=n\pi$ radians where $ n\in Z\\$$
cos^6x = cos^4x subtract cos^4x ffrom both sides
cos^6x - cos^4x = 0 faactor this
cos^4x*(cos^2x - 1) = 0
cos^4x (cosx + 1)(cosx-1) setting each factor to 0, we have
x = 0, pi/2, pi , (3 pi)/2 in the interval [0, 2pi)
More general solutions are x = 0 + (pi)n and x = pi/2 + (pi/2)n where n is an integer
Okay I missed some answers - thanks for picking that up chris but your presentation looks strange.
isn't
x = 0 + (pi)n and x = pi/2 + (pi/2)n
really just
$$x=\frac{n\pi}{2}\qquad where \qquad n\in Z$$
???