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/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2sin^2(θ)-sin(θ)=1

2 sin^{2} (θ) - sin (θ) = 1

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

2 sin^{2} (θ) - sin (θ) = 1

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

2 sin (- 3 x - \frac{π}{4}) = 2

4 sin^{3} (θ) - 3 sin (θ) = 0

\frac{sec ^{2} ( θ ) - 1}{1 - cos ^{2} ( θ )} - 1 = cot^{2} (θ)

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

4 cos (x) + 4 sin (x) tan (x) = 8

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/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin