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Answer:
- arcsin(2/3) ≈ 41.81°, 138.19°
Explanation:
Rewrite as a quadratic in sin(θ) and solve that in the usual way.
3cos(2θ) +sin(θ) = 1
3(1 -2sin²(θ)) +sin(θ) = 1 . . . . use an identity for cos(2θ)
6sin²(θ) -sin(θ) -2 = 0 . . . . . rearrange to standard form
(3sin(θ) -2)(2sin(θ) +1) = 0 . . . . factor
The values of sin(θ) that make this true are ...
sin(θ) = 2/3, sin(θ) = -1/2
In the range 0 < θ < 180°, we're only interested in ...
sin(θ) = 2/3
θ = arcsin(2/3) or 180° -arcsin(2/3)
θ ≈ {41.81°, 138.19°}