What is the general solution for 2sin^2(θ)=-3sin(θ)-1 ?

The general solution for 2sin^2(θ)=-3sin(θ)-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin

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2sin^2(θ)=-3sin(θ)-1

2 sin^{2} (θ) = - 3 sin (θ) - 1

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

2 sin^{2} (θ) = - 3 sin (θ) - 1

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin (x) = 1.9 cos (x)

tan^{2} (x) + 9 tan (x) + 1 = 0

sinh (t) = 0

9 sin (x) - 4 = 0

cos (2 x) \cdot 2 - 1 = 0

What is the general solution for 2sin^2(θ)=-3sin(θ)-1 ?
The general solution for 2sin^2(θ)=-3sin(θ)-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin