What is the general solution for 2sin^2(x)+3sin(x)+1=0 ?

The general solution for 2sin^2(x)+3sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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2sin^2(x)+3sin(x)+1=0

2 sin^{2} (x) + 3 sin (x) + 1 = 0

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

2 sin^{2} (x) + 3 sin (x) + 1 = 0

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (\frac{17 π}{6})

2 sin^{2} (x) - sin (x) - 1 = 0

cos (- \frac{π}{2})

2 cos (θ) + 2 = 0

2 sin^{2} (x) - cos (x) - 1 = 0

What is the general solution for 2sin^2(x)+3sin(x)+1=0 ?
The general solution for 2sin^2(x)+3sin(x)+1=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin