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

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

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1-cos(θ)=1

1 - cos (θ) = 1

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 - 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

csc^{2} (x) + 2 cot (x) = 1

2 cos (x) + sin (x) = 2

sin (θ) tan (θ) - sin (θ) = 0

6 cos^{2} (x) + 5 sin (x) = 7

sin (θ) = 0.9848, 0^{\circ} \leq θ < 36 0^{\circ}

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