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

The general solution for 2sin^2(x)+4sin(x)+2=0 is x=(3pi)/2+2pin

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

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

x = \frac{3 π}{2} + 2 πn

Solution steps

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

Solve by substitution

sin (x) = - 1

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

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

2 sin (θ) = csc (θ)

cos (θ) = - 0.7

2 cos^{2} (x) + 7 cos (x) + 5 = 0

2 sin (x) = - 1, 0 \leq x < 2 π

2 cos^{2} (x) = 3 - 3 sin (x)

What is the general solution for 2sin^2(x)+4sin(x)+2=0 ?
The general solution for 2sin^2(x)+4sin(x)+2=0 is x=(3pi)/2+2pin