What is the general solution for solvefor θ,3cos(θ)=1+cos(θ) ?

The general solution for solvefor θ,3cos(θ)=1+cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

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solvefor θ,3cos(θ)=1+cos(θ)

solve for

θ, 3 cos (θ) = 1 + 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

3 cos (θ) = 1 + cos (θ)

Solve by substitution

3 cos (θ) = 1 + cos (θ)

Let:

cos (θ) = u

3 u = 1 + u

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

3 u = 1 + u

Move

u

to the left side

3 u = 1 + u

Subtract

u

from both sides

3 u - u = 1 + u - u

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

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

tan (a) = \frac{5}{12}, sin (a) < 0, sin (a) + cos (a)

cot (2 x) = 3, 0 \leq x \leq 2 π

cot (θ) = \frac{5}{3}

cos (2 θ) = \frac{3}{5}

cos (\frac{x}{3}) = \frac{1}{2}

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