I can't figure out the following question:

Solve the equation for 0 less than or equal than x less than or equal than 2pi:

16cos^2x - 4cosx + 1 = 0

Aside (using "y" instead of "cos" to visualize):

16y^2 - 4y + 1 = 0

4y (4y - 1) + 1 = 0 or (4cosx[4cosx - 1] + 1 = 0)

Nevermind, I just realized you can't solve this equation