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

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

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cos^2(x)sin(x)=sin(x),(0,2pi)

cos^{2} (x) sin (x) = sin (x), (0, 2 π)

x = π

Solution steps

cos^{2} (x) sin (x) = sin (x), (0, 2 π)

Subtract

sin (x)

from both sides

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

Factor

cos^{2} (x) sin (x) - sin (x) : sin (x) (cos (x) + 1) (cos (x) - 1)

sin (x) (cos (x) + 1) (cos (x) - 1) = 0

Solving each part separately

sin (x) = 0 or cos (x) + 1 = 0 or cos (x) - 1 = 0

sin (x) = 0, (0, 2 π) : x = π

cos (x) + 1 = 0, (0, 2 π) : x = π

cos (x) - 1 = 0, (0, 2 π) :

No Solution

Combine all the solutions

x = π

cos (x) = - \frac{3}{7}

cos (x) - sin (7 x) = 0

sin (3 x) + sin (x) = 2 sin (2 x)

3 tan (\frac{1}{2} x) + 1 = 0

12 sin (θ) = 4 + 4 sin (θ)

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