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

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

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3sin(θ)+2sin^2(θ)=-1

3 sin (θ) + 2 sin^{2} (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

3 sin (θ) + 2 sin^{2} (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

cot^{2} (x) + 3 csc (x) = - 3

sec (θ) = 3

6 cos (x) = 0

sin (2 θ) - sin (4 θ) = 0

2 cos^{2} (x) + 7 cos (x) + 3 = 0

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