What is the general solution for -2sin^2(θ)+9sin(θ)=8sin(θ)-1 ?

The general solution for -2sin^2(θ)+9sin(θ)=8sin(θ)-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin

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-2sin^2(θ)+9sin(θ)=8sin(θ)-1

- 2 sin^{2} (θ) + 9 sin (θ) = 8 sin (θ) - 1

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

Solution steps

- 2 sin^{2} (θ) + 9 sin (θ) = 8 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{π}{2} + 2 πn

Combine all the solutions

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

sin^{2} (x) - sin (x) - 1 = 0

2 sec (x) - 4 = 0

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

6 tan (x) sin (x) = - 5 sin (x)

2 tan (θ) + 7 = 0

What is the general solution for -2sin^2(θ)+9sin(θ)=8sin(θ)-1 ?
The general solution for -2sin^2(θ)+9sin(θ)=8sin(θ)-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin