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

The general solution for -sin(θ)+2sin^2(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=2pin,θ=pi+2pin

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-sin(θ)+2sin^2(θ)=0

- sin (θ) + 2 sin^{2} (θ) = 0

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

Solution steps

- sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 0

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

- sec (x) + 2 sec (x) cos (x) = 0

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What is the general solution for -sin(θ)+2sin^2(θ)=0 ?
The general solution for -sin(θ)+2sin^2(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=2pin,θ=pi+2pin