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

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

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

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

x = 0, x = 2 π

Solution steps

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

General solutions for

sin (\frac{x}{2}) = 0

\frac{x}{2} = 0 + 2 πn, \frac{x}{2} = π + 2 πn

Solve

\frac{x}{2} = 0 + 2 πn : x = 4 πn

Solve

\frac{x}{2} = π + 2 πn : x = 2 π + 4 πn

x = 4 πn, x = 2 π + 4 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

sin (11 0^{\circ} - \frac{x}{2}) = - sin (7 0^{\circ})

(2 sin (x) - 1) (sin (x) + 0.5) = 0

tan (θ) - 2 cos (θ) tan (θ) = 0

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

5 cos (x) = - 2 sin^{2} (x) + 4, 0 \leq x < 2 π

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