What is the general solution for 2sin^2(x)+9sin(x)+4=0 ?

The general solution for 2sin^2(x)+9sin(x)+4=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2sin^2(x)+9sin(x)+4=0

2 sin^{2} (x) + 9 sin (x) + 4 = 0

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

Solution steps

2 sin^{2} (x) + 9 sin (x) + 4 = 0

Solve by substitution

sin (x) = - \frac{1}{2}, sin (x) = - 4

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

sin (x) = - 4 :

No Solution

Combine all the solutions

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

6 sin (2 x) = 6 cos (x)

cot (θ) = \frac{5}{12}

0 = cos^{2} (x) - sin^{2} (x)

tan (x) = \frac{7.4}{9.3}

tan^{4} (x) - 1 = 0

What is the general solution for 2sin^2(x)+9sin(x)+4=0 ?
The general solution for 2sin^2(x)+9sin(x)+4=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin