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

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

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

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

x = 0, x = π, x = 2 π

Solution steps

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

cos (x) + 1 = 0, 0 \leq x \leq 2 π : x = π

cos (x) - 1 = 0, 0 \leq x \leq 2 π : x = 0, x = 2 π

Combine all the solutions

x = 0, x = π, x = 2 π

2 \cdot cos (4 x) - 1 = 0

2 cos^{2} (θ) - 3 cos (θ) = 2

arctan (0.2 x) = 4 0^{\circ}

sin (x) - sin (2 x) + sin (3 x) = 0

sin (x) = \frac{- 2 5}{5}

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