What is the general solution for 2cos(1/5 θ)=sqrt(3) ?

The general solution for 2cos(1/5 θ)=sqrt(3) is θ=(5pi)/6+10pin,θ=(55pi)/6+10pin

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Popular >

2cos(1/5 θ)=sqrt(3)

Solution

2 cos (\frac{1}{5} θ) = 3

Solution

θ = \frac{5 π}{6} + 10 πn, θ = \frac{55 π}{6} + 10 πn

Degrees

θ = 15 0^{\circ} + 180 0^{\circ} n, θ = 165 0^{\circ} + 180 0^{\circ} n

Solution steps

2 cos (\frac{1}{5} θ) = 3

Divide both sides by

2

2 cos (\frac{1}{5} θ) = 3

Divide both sides by

2

\frac{2 cos ( \frac{1}{5} θ )}{2} = \frac{3}{2}

Simplify

cos (\frac{1}{5} θ) = \frac{3}{2}

cos (\frac{1}{5} θ) = \frac{3}{2}

General solutions for

cos (\frac{1}{5} θ) = \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{1}{5} θ = \frac{π}{6} + 2 πn, \frac{1}{5} θ = \frac{11 π}{6} + 2 πn

\frac{1}{5} θ = \frac{π}{6} + 2 πn, \frac{1}{5} θ = \frac{11 π}{6} + 2 πn

Solve

\frac{1}{5} θ = \frac{π}{6} + 2 πn : θ = \frac{5 π}{6} + 10 πn

\frac{1}{5} θ = \frac{π}{6} + 2 πn

Multiply both sides by

5

\frac{1}{5} θ = \frac{π}{6} + 2 πn

Multiply both sides by

5

5 \cdot \frac{1}{5} θ = 5 \cdot \frac{π}{6} + 5 \cdot 2 πn

Simplify

5 \cdot \frac{1}{5} θ = 5 \cdot \frac{π}{6} + 5 \cdot 2 πn

Simplify

5 \cdot \frac{1}{5} θ : θ

5 \cdot \frac{1}{5} θ

Multiply fractions:

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

= \frac{1 \cdot 5}{5} θ

Cancel the common factor:

5

= θ \cdot 1

Multiply:

θ \cdot 1 = θ

= θ

Simplify

5 \cdot \frac{π}{6} + 5 \cdot 2 πn : \frac{5 π}{6} + 10 πn

5 \cdot \frac{π}{6} + 5 \cdot 2 πn

Multiply

5 \cdot \frac{π}{6} : \frac{5 π}{6}

5 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 5}{6}

= \frac{5 π}{6} + 5 \cdot 2 πn

Multiply the numbers:

5 \cdot 2 = 10

= \frac{5 π}{6} + 10 πn

θ = \frac{5 π}{6} + 10 πn

θ = \frac{5 π}{6} + 10 πn

θ = \frac{5 π}{6} + 10 πn

Solve

\frac{1}{5} θ = \frac{11 π}{6} + 2 πn : θ = \frac{55 π}{6} + 10 πn

\frac{1}{5} θ = \frac{11 π}{6} + 2 πn

Multiply both sides by

5

\frac{1}{5} θ = \frac{11 π}{6} + 2 πn

Multiply both sides by

5

5 \cdot \frac{1}{5} θ = 5 \cdot \frac{11 π}{6} + 5 \cdot 2 πn

Simplify

5 \cdot \frac{1}{5} θ = 5 \cdot \frac{11 π}{6} + 5 \cdot 2 πn

Simplify

5 \cdot \frac{1}{5} θ : θ

5 \cdot \frac{1}{5} θ

Multiply fractions:

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

= \frac{1 \cdot 5}{5} θ

Cancel the common factor:

5

= θ \cdot 1

Multiply:

θ \cdot 1 = θ

= θ

Simplify

5 \cdot \frac{11 π}{6} + 5 \cdot 2 πn : \frac{55 π}{6} + 10 πn

5 \cdot \frac{11 π}{6} + 5 \cdot 2 πn

5 \cdot \frac{11 π}{6} = \frac{55 π}{6}

5 \cdot \frac{11 π}{6}

Multiply fractions:

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

= \frac{11 π 5}{6}

Multiply the numbers:

11 \cdot 5 = 55

= \frac{55 π}{6}

5 \cdot 2 πn = 10 πn

5 \cdot 2 πn

Multiply the numbers:

5 \cdot 2 = 10

= 10 πn

= \frac{55 π}{6} + 10 πn

θ = \frac{55 π}{6} + 10 πn

θ = \frac{55 π}{6} + 10 πn

θ = \frac{55 π}{6} + 10 πn

θ = \frac{5 π}{6} + 10 πn, θ = \frac{55 π}{6} + 10 πn

Popular Examples

sec(θ)=-7/(sqrt(13))

sec (θ) = - \frac{7}{13}

tan(x)=-infinity

tan (x) = - \infty

tan(2x)=-tan(x)2

tan (2 x) = - tan (x) 2

2sin^2(x)-5sin(x)-12=0

2 sin^{2} (x) - 5 sin (x) - 12 = 0

1/(sin(x))=sqrt(5)

\frac{1}{sin ( x )} = 5

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(1/5 θ)=sqrt(3) ?
The general solution for 2cos(1/5 θ)=sqrt(3) is θ=(5pi)/6+10pin,θ=(55pi)/6+10pin