What is the general solution for 2sin(3θ)=1 ?

The general solution for 2sin(3θ)=1 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3

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2sin(3θ)=1

2 sin (3 θ) = 1

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Solution steps

2 sin (3 θ) = 1

Divide both sides by

2

sin (3 θ) = \frac{1}{2}

General solutions for

sin (3 θ) = \frac{1}{2}

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{5 π}{6} + 2 πn

Solve

3 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

2 sin^{2} (x) + 3 sin (x) - 2 = 0

3 cos (θ) = 1 + cos (θ)

2 cos (2 θ) - 1 = 0

tan (2 x) + 3 = 0

sin^{2} (x) = 2 - 2 cos (x)

What is the general solution for 2sin(3θ)=1 ?
The general solution for 2sin(3θ)=1 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3