What is the general solution for 2sin^2(θ)=1-sin(θ) ?

The general solution for 2sin^2(θ)=1-sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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2sin^2(θ)=1-sin(θ)

2 sin^{2} (θ) = 1 - sin (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

2 sin^{2} (θ) = 1 - sin (θ)

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin (2 x) - sin (x) = 0

sin (54 0^{\circ})

arccos (0.5)

2 sin (2 x) = 1

16 cos (15 x) + 8 = 2

What is the general solution for 2sin^2(θ)=1-sin(θ) ?
The general solution for 2sin^2(θ)=1-sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin