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

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

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

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

x = π, x = 2 π

Solution steps

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

Subtract

sin (x)

from both sides

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

Factor

sin (x) cos (x) - sin (x) : sin (x) (cos (x) - 1)

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

Solving each part separately

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

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

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

Combine all the solutions

x = π, x = 2 π

cos^{5} (x) - sin (x) = 0

x, y = \frac{1}{5 sin ( 2 x )}

cos (θ) = \frac{3}{35}

csc^{2} (θ) - 3 csc (θ) + 2 = 0

sin (3 x) = 2

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