What is the general solution for 11cos(θ)=5+cos(θ) ?

The general solution for 11cos(θ)=5+cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

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11cos(θ)=5+cos(θ)

11 cos (θ) = 5 + cos (θ)

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

11 cos (θ) = 5 + cos (θ)

Solve by substitution

11 cos (θ) = 5 + cos (θ)

Let:

cos (θ) = u

11 u = 5 + u

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

11 u = 5 + u

Move

u

to the left side

11 u = 5 + u

Subtract

u

from both sides

11 u - u = 5 + u - u

Simplify

10 u = 5

10 u = 5

Divide both sides by

10

10 u = 5

Divide both sides by

10

\frac{10 u}{10} = \frac{5}{10}

Simplify

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

- 4 sin (C) - 2 = sin (C) + 1

20 (l - 2) cos (θ) - 10 = 0

cos (6 x) - cos (2 x) = 0

(2 cos (x) - 1) (cos (x) + 1) = 0

sin (θ) = - \frac{1}{5}, π < θ < \frac{2 π}{2}, cos (θ)

What is the general solution for 11cos(θ)=5+cos(θ) ?
The general solution for 11cos(θ)=5+cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin