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

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

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

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

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

Solution steps

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

Solving each part separately

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

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

tan (θ) = - \frac{3}{4}, sec (θ)

\frac{12}{sin ( 12 0 ^{\circ} )} = \frac{5}{sin ( x )}

- 3 tan^{2} (x) = - 1

3 - tan^{2} (b) = 0

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

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