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

The general solution for sin^2(θ)=8sin(θ)+9 is θ=(3pi)/2+2pin

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sin^2(θ)=8sin(θ)+9

sin^{2} (θ) = 8 sin (θ) + 9

θ = \frac{3 π}{2} + 2 πn

Solution steps

sin^{2} (θ) = 8 sin (θ) + 9

Solve by substitution

sin (θ) = 9, sin (θ) = - 1

sin (θ) = 9 :

No Solution

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

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

x, e^{c o s (y)} = cos (x)

10 sin (3 x) = 5

4 0^{2} = 2 8^{2} + 1 5^{2} - 2 (28) (15) \cdot cos (x)

5 tan^{2} (x) - 1 = 2 tan^{2} (x)

5 sin (2 x) + 8 cos (x) = 0

What is the general solution for sin^2(θ)=8sin(θ)+9 ?
The general solution for sin^2(θ)=8sin(θ)+9 is θ=(3pi)/2+2pin