What is the general solution for 5sin^2(θ)+12sin(θ)+7=0 ?

The general solution for 5sin^2(θ)+12sin(θ)+7=0 is θ=(3pi)/2+2pin

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5sin^2(θ)+12sin(θ)+7=0

5 sin^{2} (θ) + 12 sin (θ) + 7 = 0

θ = \frac{3 π}{2} + 2 πn

Solution steps

5 sin^{2} (θ) + 12 sin (θ) + 7 = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = - \frac{7}{5}

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

sin (θ) = - \frac{7}{5} :

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

3 sin (x) + 2 sin^{2} (x) = - 1

(tan (x) + 3) (2 sin (x) - 2) = 0

3 cos^{2} (x) - 2 cos (x) - 2 = 0, 0 \leq x < 2 π

sin (x + \frac{π}{4}) + sin (x - \frac{π}{4}) + \frac{6}{2} = 0

tan (θ) = \frac{500}{300}

What is the general solution for 5sin^2(θ)+12sin(θ)+7=0 ?
The general solution for 5sin^2(θ)+12sin(θ)+7=0 is θ=(3pi)/2+2pin