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= pi/2 ,x=0,x=pi,x=2pi

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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 = \frac{π}{2}, x = 0, x = π, x = 2 π

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

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

sin (x + 3 7^{\circ}) = cos (2 x + 8)

sin (θ) = \frac{8}{17}, cos (θ) = \frac{15}{17}

tan (x) = 4746050429

cos^{2} (5 x) = 1

1 = 1 - 2 arctan (\frac{1 - 2 - C}{1 - C})

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= pi/2 ,x=0,x=pi,x=2pi