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

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

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2=(1-cos(θ))

2 = (1 - cos (θ))

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 = (1 - cos (θ))

Switch sides

(1 - cos (θ)) = 2

1 - cos (θ) = 2

Move

1

to the right side

1 - cos (θ) = 2

Subtract

1

from both sides

1 - cos (θ) - 1 = 2 - 1

Simplify

- cos (θ) = 1

- cos (θ) = 1

Divide both sides by

- 1

- cos (θ) = 1

Divide both sides by

- 1

\frac{- cos ( θ )}{- 1} = \frac{1}{- 1}

Simplify

cos (θ) = - 1

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

sin (120 \cdot π \cdot t) = 0.06141

cos (θ) = \frac{2 13}{13}

cos^{2} (A) + 2 = 3 cos (A)

sin (θ) = \frac{- 5}{6}

\frac{sin ( x )}{16} = \frac{sin ( 3 1 ^{\circ} )}{12}

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