What is the general solution for 2+cos(θ)=5cos(θ) ?

The general solution for 2+cos(θ)=5cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2+cos(θ)=5cos(θ)

Solution

2 + cos (θ) = 5 cos (θ)

Solution

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

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 + cos (θ) = 5 cos (θ)

Solve by substitution

2 + cos (θ) = 5 cos (θ)

Let:

cos (θ) = u

2 + u = 5 u

2 + u = 5 u : u = \frac{1}{2}

2 + u = 5 u

Move

2

to the right side

2 + u = 5 u

Subtract

2

from both sides

2 + u - 2 = 5 u - 2

Simplify

u = 5 u - 2

u = 5 u - 2

Move

5 u

to the left side

u = 5 u - 2

Subtract

5 u

from both sides

u - 5 u = 5 u - 2 - 5 u

Simplify

- 4 u = - 2

- 4 u = - 2

Divide both sides by

- 4

- 4 u = - 2

Divide both sides by

- 4

\frac{- 4 u}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 u}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 u}{- 4} : u

\frac{- 4 u}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4 u}{4}

Divide the numbers:

\frac{4}{4} = 1

= u

Simplify

\frac{- 2}{- 4} : \frac{1}{2}

\frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{1}{2}

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{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}

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

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

Combine all the solutions

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

Popular Examples

11=12+3.1cos((2pi)/(365)t)

11 = 12 + 3.1 cos (\frac{2 π}{365} t)

solvefor θ,6sin^2(θ)-3=0 solve for

θ, 6 sin^{2} (θ) - 3 = 0

sin(x)=-7/8

sin (x) = - \frac{7}{8}

sin(x)=0.942

sin (x) = 0.942

4cos^2(x)=5+4sin(x),(0,2pi)

4 cos^{2} (x) = 5 + 4 sin (x), (0, 2 π)

Frequently Asked Questions (FAQ)

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