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

The general solution for sin^2(x)+2sin(x)+1=0,0<= x<2pi is x=(3pi)/2

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sin^2(x)+2sin(x)+1=0,0<= x<2pi

sin^{2} (x) + 2 sin (x) + 1 = 0, 0 \leq x < 2 π

x = \frac{3 π}{2}

Solution steps

sin^{2} (x) + 2 sin (x) + 1 = 0, 0 \leq x < 2 π

Solve by substitution

sin (x) = - 1

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{3 π}{2}

2 + 7 cos (θ) = 4 sin^{2} (θ)

x, sinh (x) = 0

sin (a) = 0.5

2 sin^{4} (x) = cos^{2} (x)

2 cos (3 x) = cot (3 x)

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