SOLUTION: Solve for 0≤theta≤360, 2cos^2 theta +5cos theta-3=0

 = 0,

Introduce new variable x = . Then your equation takes the form

 = 0.

Apply the quadratic formula:

 =  = .

 = ,   = -3.

The second root is not acceptable, since cosine must be <= 1 by modulus.

The only solution is  = ,

which gives  = 60 degs and/or   = 300 degs.

Answer.   = 60 degs  and/or   = 300 degs.