What is the general solution for 7sin^2(θ)+15sin(θ)+8=0 ?

The general solution for 7sin^2(θ)+15sin(θ)+8=0 is θ=(3pi)/2+2pin

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7sin^2(θ)+15sin(θ)+8=0

7 sin^{2} (θ) + 15 sin (θ) + 8 = 0

θ = \frac{3 π}{2} + 2 πn

Solution steps

7 sin^{2} (θ) + 15 sin (θ) + 8 = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = - \frac{8}{7}

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

sin (θ) = - \frac{8}{7} :

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

2 sin^{23} (x) + sin^{26} (x) - 2 = 0

arccos (3 x) = 1

sin (θ) - cos (θ) = \frac{3}{2}, θ [\frac{3 π}{2}, 2 π]

1 6^{s i n (x)} = 8^{c s c (x)}

tan^{2} (x) - 6 tan (x) - 7 = 0

What is the general solution for 7sin^2(θ)+15sin(θ)+8=0 ?
The general solution for 7sin^2(θ)+15sin(θ)+8=0 is θ=(3pi)/2+2pin