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

The general solution for 2sin^2(x)-sin(x)=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6 ,x=0,x=pi,x=2pi

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}, x = 0, x = π, x = 2 π

Solution steps

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

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

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

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = 0, x = π, x = 2 π

tan (x) = \frac{2}{2}

4 cos (2 x) - 3 = 0

sin^{2} (x) + 1 = 0

cos (t) - sin (2 t) = 0

sin (2 x - \frac{π}{2}) = - 1

What is the general solution for 2sin^2(x)-sin(x)=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)-sin(x)=0,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6 ,x=0,x=pi,x=2pi