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

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

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1=1-cos(θ)

1 = 1 - cos (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

1 = 1 - cos (θ)

Switch sides

1 - cos (θ) = 1

Subtract

1

from both sides

1 - cos (θ) - 1 = 1 - 1

Simplify

- cos (θ) = 0

Divide both sides by

- 1

- cos (θ) = 0

Divide both sides by

- 1

\frac{- cos ( θ )}{- 1} = \frac{0}{- 1}

Simplify

cos (θ) = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

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}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

tan (\frac{θ}{3}) = 3 sin (4 8^{\circ})

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} = csc (x)

2 cos^{2} (x) - 9 sin (x) - 3 = 0

3 + \frac{1}{5} \cdot cos (\frac{θ}{3} + \frac{7 π}{4}) = 3

cosh (2 x) = 2 cosh (x) - 1

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