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

The general solution for sin(2x)cos(x)=0,0<= x<= 2pi is x=0,x= pi/2 ,x=pi,x=(3pi)/2 ,x=2pi

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

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

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

Solution steps

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

Solving each part separately

sin (2 x) = 0 or cos (x) = 0

sin (2 x) = 0, 0 \leq x \leq 2 π : x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

Combine all the solutions

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

tan (x + π) + 2 sin (x + π) = 0

sin (x) = \frac{3}{2}

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

sin (θ) - 1 = cos (θ)

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

What is the general solution for sin(2x)cos(x)=0,0<= x<= 2pi ?
The general solution for sin(2x)cos(x)=0,0<= x<= 2pi is x=0,x= pi/2 ,x=pi,x=(3pi)/2 ,x=2pi