What is the general solution for 2sin^2(θ)+13sin(θ)+6=0 ?

The general solution for 2sin^2(θ)+13sin(θ)+6=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2sin^2(θ)+13sin(θ)+6=0

2 sin^{2} (θ) + 13 sin (θ) + 6 = 0

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

2 sin^{2} (θ) + 13 sin (θ) + 6 = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = - 6

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - 6 :

No Solution

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

- 3 tan (- x) = 5 tan (x) - 3

tan (a) = 22

sin (2 x) = 3 cos (x)

cos (2 x) = 685

3 sin (C) - 10 = 5 sin (C) - 9

What is the general solution for 2sin^2(θ)+13sin(θ)+6=0 ?
The general solution for 2sin^2(θ)+13sin(θ)+6=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin