What is the general solution for sin(x/2)=1,0<= x<2pi ?

The general solution for sin(x/2)=1,0<= x<2pi is x=pi

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sin(x/2)=1,0<= x<2pi

sin (\frac{x}{2}) = 1, 0 \leq x < 2 π

x = π

Solution steps

sin (\frac{x}{2}) = 1, 0 \leq x < 2 π

General solutions for

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

\frac{x}{2} = \frac{π}{2} + 2 πn

Solve

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

x = π + 4 πn

Solutions for the range

0 \leq x < 2 π

x = π

sin (\frac{x}{2}) = - cos (x - \frac{π}{3})

3 = cot^{2} (θ) + 4 cot (θ)

cos^{2} (2 x) + 3 sin (2 x) - 3 = 0

1 - cos^{2} (x) = 0.75 1 - cos^{2} (y)

sec^{2} (x) - 6 tan (x) + 4 = 0

What is the general solution for sin(x/2)=1,0<= x<2pi ?
The general solution for sin(x/2)=1,0<= x<2pi is x=pi