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

Degrees

x = 18 0^{\circ}

Solution steps

sin (\frac{x}{2}) = 1, 0 \leq x < 2 π

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{x}{2} = \frac{π}{2} + 2 πn

\frac{x}{2} = \frac{π}{2} + 2 πn

Solve

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

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{π}{2} + 2 \cdot 2 πn : π + 4 πn

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 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