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

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

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1-cos(θ)=1+cos(θ)

1 - cos (θ) = 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 - cos (θ) = 1 + cos (θ)

Solve by substitution

1 - cos (θ) = 1 + cos (θ)

Let:

cos (θ) = u

1 - u = 1 + u

1 - u = 1 + u : u = 0

1 - u = 1 + u

Subtract

1

from both sides

1 - u - 1 = 1 + u - 1

Simplify

- u = u

Move

u

to the left side

- u = u

Subtract

u

from both sides

- u - u = u - u

Simplify

- 2 u = 0

- 2 u = 0

Divide both sides by

- 2

- 2 u = 0

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{0}{- 2}

Simplify

u = 0

u = 0

Substitute back

u = cos (θ)

cos (θ) = 0

cos (θ) = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

cos (2 x) + cos (x) = 1

tan (x) sec (x) = - 2 tan (x)

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

sin (\frac{3 π}{2} + x) + sin (\frac{3 π}{2} + x) = - 2

cos (x) tan (x) = 0

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