What is the general solution for sin(θ)(2sin(θ)+1)=0 ?

The general solution for sin(θ)(2sin(θ)+1)=0 is θ=2pin,θ=pi+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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sin(θ)(2sin(θ)+1)=0

sin (θ) (2 sin (θ) + 1) = 0

θ = 2 πn, θ = π + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

sin (θ) (2 sin (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 sin (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

2 sin (θ) + 1 = 0 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

tan (θ) = 0.437

sin (\frac{x}{3}) = - \frac{3}{2}

2 cos (x) + 3 = 2

6 sin^{2} (x) = 3

tan (x) cot (x) = csc (x) sec (x)

What is the general solution for sin(θ)(2sin(θ)+1)=0 ?
The general solution for sin(θ)(2sin(θ)+1)=0 is θ=2pin,θ=pi+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin