What is the general solution for 1-cos(θ)=0 ?

The general solution for 1-cos(θ)=0 is θ=2pin

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1-cos(θ)=0

1 - cos (θ) = 0

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - cos (θ) = 0

Move

1

to the right side

1 - cos (θ) = 0

Subtract

1

from both sides

1 - cos (θ) - 1 = 0 - 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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sin^{2} (x) + sin^{2} (2 x) = 1

tan (2 x - 5) = 1

\frac{5}{21 \cdot 30} = cos (x)

cos^{2} (x) = \frac{( 7 k - 7 )}{3}

tan (x) - 2 sin (x) cos (x) = 0

What is the general solution for 1-cos(θ)=0 ?
The general solution for 1-cos(θ)=0 is θ=2pin