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Answer:
x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}
Explanation:
Using the trig identity ...
cos(2x) = 1 -2sin²(x)
we can rewrite the equation as ...
(1 -2sin²(3x)) -5sin(3x) +2 = 0
2sin²(3x) +5sin(3x) -3 = 0 . . . . . standard form
(sin(3x) +3)(2sin(3x) -1) = 0 . . . . factored form
There are no real values of x that will make the first factor zero. The second factor is zero when ...
sin(3x) = 1/2
3x = arcsin(1/2) +2nπ or (2n+1)π -arcsin(1/2)
x = 1/3(π/6 +2nπ) or 1/3((2n +1)π -π/6)
x = (12n+1)π/18 or (12n+5)π/18
x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}