Question

10 th
3tan^2 (2theta) - 1)(2sin 3theta - 1) = 0 find theta
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Answer 1

Answer:

3⋅

cos

2

θ

sin

2

θ

−2sinθ=0,cosθ



=0.

Multiplying throughout by cos

2

θ and then changing it into sin

2

θ. we get

3sin

2

θ−2sinθ(1−sin

2

θ)=0

sinθ(3sinθ−2+2sin

2

θ)=0

sinθ(2sin

2

θ+4sinθ−sinθ−2)=0

sinθ(sinθ+2)(2sinθ−1)=0

∴sinθ=0, ∴θ=nπ

sinθ=1/2=sin(π/6),

∴θ=nπ+(−1)

n

π/6

sinθ=−2(rejected as sinθ cannot be greater than one numerically).