What is the general solution for 6sin^2(θ)+13sin(θ)+7=0 ?

The general solution for 6sin^2(θ)+13sin(θ)+7=0 is θ=(3pi)/2+2pin

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6sin^2(θ)+13sin(θ)+7=0

6 sin^{2} (θ) + 13 sin (θ) + 7 = 0

θ = \frac{3 π}{2} + 2 πn

Solution steps

6 sin^{2} (θ) + 13 sin (θ) + 7 = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = - \frac{7}{6}

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

sin (θ) = - \frac{7}{6} :

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

12 sin^{2} (x) - 17 sin (x) + 6 = 0

sin (θ) = \frac{3}{3}

tan (x + π) + cos (x + \frac{π}{2}) = 0

sin (θ) cos (θ) = \frac{3}{4}

tan (x + \frac{π}{6}) = 3

What is the general solution for 6sin^2(θ)+13sin(θ)+7=0 ?
The general solution for 6sin^2(θ)+13sin(θ)+7=0 is θ=(3pi)/2+2pin