What is the general solution for 2sin^2(θ)+9sin(θ)-5=0 ?

The general solution for 2sin^2(θ)+9sin(θ)-5=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2sin^2(θ)+9sin(θ)-5=0

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 5

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - 5 :

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (x) tan (x) - sin (x) = 0

sinh (x) = \frac{15}{8}

(2 cos (x) - 1) (2 sin (x) - 3) = 0

sin (α) + cos (α) = \frac{1}{2}

2 sin^{2} (θ) - cos (2 θ) = 1

What is the general solution for 2sin^2(θ)+9sin(θ)-5=0 ?
The general solution for 2sin^2(θ)+9sin(θ)-5=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin