What is the general solution for 5sin^2(θ)+11sin(θ)+6=0 ?

The general solution for 5sin^2(θ)+11sin(θ)+6=0 is θ=(3pi)/2+2pin

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5sin^2(θ)+11sin(θ)+6=0

5 sin^{2} (θ) + 11 sin (θ) + 6 = 0

θ = \frac{3 π}{2} + 2 πn

Solution steps

5 sin^{2} (θ) + 11 sin (θ) + 6 = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = - \frac{6}{5}

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

sin (θ) = - \frac{6}{5} :

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

cot^{2} (x) - 4 cot (x) + 3 = 0

2 cos (x) + 1 = sec (x)

2 sin (θ) = 1, 0 \leq θ \leq 2 π

sin (8 θ) - sin (6 θ) = cos (7 θ), 0 \leq θ \leq 2 π

cos (2 x) + 14 sin^{2} (x) = 10

What is the general solution for 5sin^2(θ)+11sin(θ)+6=0 ?
The general solution for 5sin^2(θ)+11sin(θ)+6=0 is θ=(3pi)/2+2pin