Question 613354
Find all values of x to the nearest degree that satisfy the equation 3cos2x + cosx + 2 = 0 if  0o < x < 360
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Using the identity cos(2x) = 2cos^2(x) - 1, we can write the equation as
3(2cos^2(x)-1) + cos(x) + 2 = 0
Simplifying gives
6cos^2(x) + cos(x) - 1 = 0
This can be factored as
(3cos(x)-1)(2cos(x)+1) = 0
This gives cos(x) = 1/3 and cos(x) = -1/2
The 1st 2 angles are 71 and 120 deg.  
I'll leave it as an exercise to figure out the rest (there are 2 more).