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

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

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solvefor θ,2sin^2(θ)=1-sin(θ)

solve for

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

cos (a) = \frac{57}{113 \cdot 78}

tan^{2} (x) + tan (x) - 12 = 2

x, cos^{2} (x + x y) = \frac{3}{2}

sin (o) = \frac{3}{5}

a cos (x) = cos (x)

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