What is the general solution for cos(θ)=1+2cos(θ) ?

The general solution for cos(θ)=1+2cos(θ) is θ=pi+2pin

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cos(θ)=1+2cos(θ)

cos (θ) = 1 + 2 cos (θ)

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (θ) = 1 + 2 cos (θ)

Solve by substitution

cos (θ) = 1 + 2 cos (θ)

Let:

cos (θ) = u

u = 1 + 2 u

u = 1 + 2 u : u = - 1

u = 1 + 2 u

Move

2 u

to the left side

u = 1 + 2 u

Subtract

2 u

from both sides

u - 2 u = 1 + 2 u - 2 u

Simplify

- u = 1

- u = 1

Divide both sides by

- 1

- u = 1

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{1}{- 1}

Simplify

u = - 1

u = - 1

Substitute back

u = cos (θ)

cos (θ) = - 1

cos (θ) = - 1

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = π + 2 πn

cos (β) = \frac{2}{7}

tan (x) = \frac{930}{1793}

sin (x) + cos (x) = 0.2

sin (x) cos (x) = - \frac{2}{4}

sin (a) = 0.3

What is the general solution for cos(θ)=1+2cos(θ) ?
The general solution for cos(θ)=1+2cos(θ) is θ=pi+2pin