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3cos^2 thetha+2cos theta = 1 interval [0deg,360deg)
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The answer presented in the post by @MathLover1 is absurdist and incorrect.
I came to bring you a correct answer together with a correct solution.
Let cos(theta) = u. Then the original equation is
3u^2 + 2u = 1.
Transform it equivalently
3u^2 + 2u - 1 = 0,
(3u-1)*(u+1) = 0 <<<---=== by factoring.
So, the quadratic equation has two solutions: u= -1 and u= 1/3.
If u= -1, it means = -1, hence, = 180 degrees, in the assigned interval.
If u= 1/3, it means = 1/3. Then there are two solutions for in the assigned interval
= arccos(1/3) = 70.53 degrees and = 360-70.53 = 289.47 degrees.
ANSWER. The solutions for in the assigned interval are 180 degs, 70.53 degs and 289.47 degs.
Solved.