What is the general solution for sin^2(x)+8sin(x)-9=0 ?

The general solution for sin^2(x)+8sin(x)-9=0 is x= pi/2+2pin

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sin^2(x)+8sin(x)-9=0

sin^{2} (x) + 8 sin (x) - 9 = 0

x = \frac{π}{2} + 2 πn

Solution steps

sin^{2} (x) + 8 sin (x) - 9 = 0

Solve by substitution

sin (x) = 1, sin (x) = - 9

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - 9 :

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

cot^{2} (θ) + 4 csc (θ) = - 5

10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

2 cos^{2} (θ) + 7 sin (θ) = 5

\frac{2 3}{3} = sec (θ)

What is the general solution for sin^2(x)+8sin(x)-9=0 ?
The general solution for sin^2(x)+8sin(x)-9=0 is x= pi/2+2pin