What is the general solution for sin^2(x/2)-1=0 ?

The general solution for sin^2(x/2)-1=0 is x=pi+4pin,x=3pi+4pin

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sin^2(x/2)-1=0

Solution

sin^{2} (\frac{x}{2}) - 1 = 0

Solution

x = π + 4 πn, x = 3 π + 4 πn

Degrees

x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n

Solution steps

sin^{2} (\frac{x}{2}) - 1 = 0

Solve by substitution

sin^{2} (\frac{x}{2}) - 1 = 0

Let:

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

u^{2} - 1 = 0

u^{2} - 1 = 0 : u = 1, u = - 1

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

u^{2} - 1 + 1 = 0 + 1

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

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

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

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

sin (\frac{x}{2}) = 1 : x = π + 4 πn

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

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

sin (\frac{x}{2}) = - 1 : x = 3 π + 4 πn

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

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{3 π}{2} + 2 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn

Solve

\frac{x}{2} = \frac{3 π}{2} + 2 πn : x = 3 π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{3 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{3 π}{2} + 2 \cdot 2 πn : 3 π + 4 πn

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

Multiply fractions:

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

= \frac{3 π 2}{2}

Cancel the common factor:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

Combine all the solutions

x = π + 4 πn, x = 3 π + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x/2)-1=0 ?
The general solution for sin^2(x/2)-1=0 is x=pi+4pin,x=3pi+4pin