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

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

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2sin^2(θ)=sin(θ)

2 sin^{2} (θ) = sin (θ)

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

Solution steps

2 sin^{2} (θ) = sin (θ)

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

3 sin (x) = 4 cos (x)

csc^{2} (x) - csc (x) + 9 = 11

4 sin (x) = - cos^{2} (x) + 4

csc^{2} (x) + csc (x) = 2

cos (t) = - 1

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