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

The general solution for cos(θ/2)=(sqrt(3))/2 is θ= pi/3+4pin,θ=(11pi)/3+4pin

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cos(θ/2)=(sqrt(3))/2

Solution

cos (\frac{θ}{2}) = \frac{3}{2}

Solution

θ = \frac{π}{3} + 4 πn, θ = \frac{11 π}{3} + 4 πn

Degrees

θ = 6 0^{\circ} + 72 0^{\circ} n, θ = 66 0^{\circ} + 72 0^{\circ} n

Solution steps

cos (\frac{θ}{2}) = \frac{3}{2}

General solutions for

cos (\frac{θ}{2}) = \frac{3}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{θ}{2} = \frac{π}{6} + 2 πn, \frac{θ}{2} = \frac{11 π}{6} + 2 πn

\frac{θ}{2} = \frac{π}{6} + 2 πn, \frac{θ}{2} = \frac{11 π}{6} + 2 πn

Solve

\frac{θ}{2} = \frac{π}{6} + 2 πn : θ = \frac{π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 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{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

Solve

\frac{θ}{2} = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{3} + 4 πn

\frac{θ}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{11 π}{6} + 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{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

Multiply fractions:

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

= \frac{11 π 2}{6}

Multiply the numbers:

11 \cdot 2 = 22

= \frac{22 π}{6}

Cancel the common factor:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn, θ = \frac{11 π}{3} + 4 πn

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sin(a)= 1/2

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

sin^2(x)-cos^2(x)=1

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ/2)=(sqrt(3))/2 ?
The general solution for cos(θ/2)=(sqrt(3))/2 is θ= pi/3+4pin,θ=(11pi)/3+4pin