0=2cos(2x)-cos(x)....find x?

cos(2x) = 2cos²(x) - 1. So

2cos(2x) - cos(x) = 4cos²(x) - 2 - cos(x) = 0 ==>

cos(x) = 1/8 ± √(33)/8

Both of these numbers are between -1 and 1. To get all possible solutions

x = ± arccos((1 + √(33))/8) + 2nπ or x = ± arccos((1 - √(33))/8) + 2nπ

n any integer.

The first representation gives all quadrant I and IV solution and the second gives all quadrant II and III solutions....Show more

0=2[cos^2(x) - sin^2(x)]-cos(x)
0=2cos^2(x)-2sin^2(x)-cos(x)
0=2cos^2(x) - 2[1-cos^2(x)] -cos(x)
0=4cos^2(x)-cos(x)-2
Use the quadratic formula
cos(x) = 0.843 or -0.593
x = 32.54, 327.46, 126.37 and 233.63 degrees...Show more

http://www.wolframalpha.com/input/?i=0%3D2cos%282x%29-cos%28x%29...Show more