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

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

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

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

x = 0, x = π

Solution steps

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

Solve by substitution

sin (x) = 2, sin (x) = 0

sin (x) = 2, 0 \leq x < 2 π :

No Solution

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

Combine all the solutions

x = 0, x = π

cos (2 x) = \frac{( 5 k - 8 )}{5}

x, - f = sin (x)

\frac{1}{2} cos (2 x - \frac{π}{3}) + 4 = 0

sin (B) = \frac{16 sin ( 10 4 ^{\circ} )}{25}

sin (t) = \frac{3}{5}, cos (t) = \frac{4}{5}

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