What is the general solution for sin(θ/2)=-(sqrt(3))/2 ?

The general solution for sin(θ/2)=-(sqrt(3))/2 is θ=(8pi)/3+4pin,θ=(10pi)/3+4pin

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sin(θ/2)=-(sqrt(3))/2

Solution

sin (\frac{θ}{2}) = - \frac{3}{2}

Solution

θ = \frac{8 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Degrees

θ = 48 0^{\circ} + 72 0^{\circ} n, θ = 60 0^{\circ} + 72 0^{\circ} n

Solution steps

sin (\frac{θ}{2}) = - \frac{3}{2}

General solutions for

sin (\frac{θ}{2}) = - \frac{3}{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{4 π}{3} + 2 πn, \frac{θ}{2} = \frac{5 π}{3} + 2 πn

\frac{θ}{2} = \frac{4 π}{3} + 2 πn, \frac{θ}{2} = \frac{5 π}{3} + 2 πn

Solve

\frac{θ}{2} = \frac{4 π}{3} + 2 πn : θ = \frac{8 π}{3} + 4 πn

\frac{θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{4 π}{3} + 2 πn

Multiply both sides by

2

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

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{4 π}{3} + 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{4 π}{3} + 2 \cdot 2 πn : \frac{8 π}{3} + 4 πn

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{4 π}{3} = \frac{8 π}{3}

2 \cdot \frac{4 π}{3}

Multiply fractions:

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

= \frac{4 π 2}{3}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{8 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn

Solve

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 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}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

Multiply fractions:

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

= \frac{5 π 2}{3}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{8 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Popular Examples

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cos (θ) = 0.98

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sec^{2} (x) = 3 tan^{2} (x) - 1

csc(x)+cot(x)=(sqrt(3))/3

csc (x) + cot (x) = \frac{3}{3}

sin(θ)+1=cos(θ)

sin (θ) + 1 = cos (θ)

sqrt(2)sin^2(θ)-sin(θ)=0

2 sin^{2} (θ) - sin (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(θ/2)=-(sqrt(3))/2 ?
The general solution for sin(θ/2)=-(sqrt(3))/2 is θ=(8pi)/3+4pin,θ=(10pi)/3+4pin