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

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

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sin(θ)+2sin^2(θ)=0

sin (θ) + 2 sin^{2} (θ) = 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^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = - \frac{1}{2} : 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

(cot (x) + 1) (sin (x) + 1) = 0

cos^{2} (x) - 2 sin (x) = 0

sin (x) cos (2 x) + cos (x) sin (2 x) = \frac{1}{2}

sin (3 x) - 3 sin (x) = 0

sin (x) = \frac{2.1}{4}

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