Solve 2cosθ+1=0 for -180°≤θ≤180°

2cosθ + 1 = 0
2cosθ = -1
cosθ = -1/2

first find cosθ" = 1/2
θ" = arccos(1/2)
θ" = 60 degrees

but, cosθ must be negative

cosθ is negative in the 2nd and 3rd quadrants, which correspond to θ = 180 - θ" and θ = 180 + θ"
therefore,
θ = 180 - 60
= 120
θ = 180 + 60
= 240

but, last value doesn't fit into -180°≤θ≤180°
240 can also be written as - 120 degrees which does fit in

therefore,

θ = 120 and -120 degrees

I think this one is the easiest trigonometrical equation I have ever solved,because only side change will give the solution,
Since I can not write Theta,I am using A
cosA=-1/2,or cosA=cos120,or A=360n(+or-)120.Now put optimum values of n (n is not irrational or fraction) to get solutions within your limit.

cosθ = -1/2
θ = 120°