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

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

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cos(θ)=1-cos(θ)

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

cos (θ) = 1 - cos (θ)

Solve by substitution

cos (θ) = 1 - cos (θ)

Let:

cos (θ) = u

u = 1 - u

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

u = 1 - u

Move

u

to the left side

u = 1 - u

Add

u

to both sides

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 (θ) = - 1.5

tan (θ) = \frac{30}{20}

10 cos (θ) - 53 = 0

tan (θ) = \frac{8}{- 6}

tan (θ) = - 0.5

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