Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0.

Suggestion: recall that cos2x can be rewriiten as one of the following:

cos^2x - sin^2x, or
1-2sin^2x, or
2cos^2x - 1

Note that if you substitute the 3rd version into the original equation you can recast the original equation as a quadratic, meaning in the form Acos^2x + Bcos^x + C = 0. Then you can factor this to find values of cosx, and from that determine values for x. You should find that there are 2 values of cos(x), and hence 4 possible solutions for x.