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

Add

u

to 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

4 cos (x) - 22 = 0

sin (x) cos (3 x) + cos (x) sin (3 x) = 0

4 sin^{2} (x) + 4 = 5

2 csc (θ) + 3 = 0

2 cos (x) - 2 sin (2 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