What is the general solution for 2sin^2(θ)+3sin(θ)+1=0 ?

The general solution for 2sin^2(θ)+3sin(θ)+1=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin

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2sin^2(θ)+3sin(θ)+1=0

2 sin^{2} (θ) + 3 sin (θ) + 1 = 0

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 = 0

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

1 = sin (x)

sin (2 x) = - cos (x)

cos (2 θ) = cos (θ)

2 - 4 sin^{2} (x) = 0

tan (θ) sin (θ) - 4 tan (θ) = - 3 tan (θ)

What is the general solution for 2sin^2(θ)+3sin(θ)+1=0 ?
The general solution for 2sin^2(θ)+3sin(θ)+1=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin