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