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

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

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2sin^2(θ)+sin(θ)=1

2 sin^{2} (θ) + sin (θ) = 1

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) + sin (θ) = 1

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

cot^{2} (x) + csc (x) = 1

sin (x) = \frac{- 2}{2}

sin (x) = - \frac{1}{2}

cos (2 x) - 2 cos (x) = - 1

cos (x) = \frac{3}{2}

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