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

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

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

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

x = 0, x = π, x = 2 π, x = \frac{3 π}{2}

Solution steps

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

Solve by substitution

sin (x) = 0, sin (x) = - 1

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

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

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{3 π}{2}

\frac{1}{tan ( x )} = - 1

64 cosh^{4} (x) - 64 cosh^{2} (x) - 9 = 0

1 + cos (θ) = 1

tan (2 θ) = 0.4

csc (\frac{θ}{2}) = sin (\frac{θ}{2})

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