What is the general solution for sin(θ/2)=-1/(sqrt(2)) ?

The general solution for sin(θ/2)=-1/(sqrt(2)) is θ=(5pi)/2+4pin,θ=(7pi)/2+4pin

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sin(θ/2)=-1/(sqrt(2))

Solution

sin (\frac{θ}{2}) = - \frac{1}{2}

Solution

θ = \frac{5 π}{2} + 4 πn, θ = \frac{7 π}{2} + 4 πn

Solution steps

sin (\frac{θ}{2}) = - \frac{1}{2}

Simplify

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

sin (\frac{θ}{2}) = - \frac{2}{2}

General solutions for

sin (\frac{θ}{2}) = - \frac{2}{2}

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{θ}{2} = \frac{5 π}{4} + 2 πn, \frac{θ}{2} = \frac{7 π}{4} + 2 πn

\frac{θ}{2} = \frac{5 π}{4} + 2 πn, \frac{θ}{2} = \frac{7 π}{4} + 2 πn

Solve

\frac{θ}{2} = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{2} + 4 πn

\frac{θ}{2} = \frac{5 π}{4} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{4} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn : \frac{5 π}{2} + 4 πn

2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{4} = \frac{5 π}{2}

2 \cdot \frac{5 π}{4}

Multiply fractions:

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

= \frac{5 π 2}{4}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{4}

Cancel the common factor:

2

= \frac{5 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

Solve

\frac{θ}{2} = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{2} + 4 πn

\frac{θ}{2} = \frac{7 π}{4} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{7 π}{4} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn : \frac{7 π}{2} + 4 πn

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{4} = \frac{7 π}{2}

2 \cdot \frac{7 π}{4}

Multiply fractions:

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

= \frac{7 π 2}{4}

Multiply the numbers:

7 \cdot 2 = 14

= \frac{14 π}{4}

Cancel the common factor:

2

= \frac{7 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn, θ = \frac{7 π}{2} + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ/2)=-1/(sqrt(2)) ?
The general solution for sin(θ/2)=-1/(sqrt(2)) is θ=(5pi)/2+4pin,θ=(7pi)/2+4pin