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Answer:
x = {2π/3, 4π/3}
Explanation:
The equation can be factored as ...
(cos(x) +3)(2cos(x) +1) = 0
The solutions are the values of x where the factors are zero.
cos(x) +3 = 0
cos(x) = -3 . . . . . . no real solution
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2cos(x) +1 = 0
cos(x) = -1/2
x = arccos(-1/2) = 2π/3
The other solution in the desired interval is ...
x = 2π -(2π/3) = 4π/3
The solutions are x∈ {2π/3, 4π/3}.