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

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

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

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

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

Solution steps

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

Solve by substitution

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

3 sin^{2} (t) - 2 sin (t) - 1 = 0

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

tan (2 x) = 4

2 sin (2 x) = 1, in t er v a l (0, 36 0^{\circ})

sinh (x + 1) = 0

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