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

The general solution for sin^2(θ)+sin(θ)=0,0<= θ<2pi is θ=0,θ=pi,θ=(3pi)/2

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sin^2(θ)+sin(θ)=0,0<= θ<2pi

sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

θ = 0, θ = π, θ = \frac{3 π}{2}

Solution steps

sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

Solve by substitution

sin (θ) = 0, sin (θ) = - 1

sin (θ) = 0, 0 \leq θ < 2 π : θ = 0, θ = π

sin (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{2}

Combine all the solutions

θ = 0, θ = π, θ = \frac{3 π}{2}

16 tan^{2} (θ) - 1 = 0

sin^{2} (x) + \frac{5}{14} = \frac{6}{7}

sin (2 θ) - 2 sin (θ) = 0

3 sin (t) + cos (t) = 1

cos^{2} (x) = cos (x) + 1

What is the general solution for sin^2(θ)+sin(θ)=0,0<= θ<2pi ?
The general solution for sin^2(θ)+sin(θ)=0,0<= θ<2pi is θ=0,θ=pi,θ=(3pi)/2