Can someone help me solve this basic trig problem it's really easy but I don't remember how 2cos^2 x =cosx?

2cos^2x = cosx

Subtract cosx from both sides to get it equal to zero

2cos^2x - cosx = 0

Now factor the left

cosx(2cosx - 1) = 0

Set each factor equal to zero.

2cosx - 1 = 0 | cosx = 0

cosx = 1/2 | cosx = 0

the reference angle for the first is 60 degrees. The corresponding angle in the fourth quadrant with this reference angle is 300 degrees. The second is a quadrantal angle 90 and 270 degrees.

i'm unsure if it relatively is sweet or no longer, i do no longer relatively comprehend Trig all that nicely, yet... pi/4 is comparable to 40 5 stages as a reference perspective on the Unit Circle that's the sq. Root of two/ 2 returned, unsure if it relatively is what your searching for yet i wish it facilitates somewhat....