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

The general solution for 1+cos(θ)=-cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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1+cos(θ)=-cos(θ)

1 + cos (θ) = - cos (θ)

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos (θ) = - cos (θ)

Solve by substitution

1 + cos (θ) = - cos (θ)

Let:

cos (θ) = u

1 + u = - u

1 + u = - u : u = - \frac{1}{2}

1 + u = - u

Move

1

to the right side

1 + u = - u

Subtract

1

from both sides

1 + u - 1 = - u - 1

Simplify

u = - u - 1

u = - u - 1

Move

u

to the left side

u = - u - 1

Add

u

to both sides

u + u = - u - 1 + u

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

\frac{2 u}{2} = \frac{- 1}{2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

2 cos^{2} (x) - cos (- x) - 1 = 0

3 sec^{2} (u) + 7 tan (u) = 3

3 sin (t) = 2 cos^{2} (t)

sin (θ) = \frac{12 sin ( 4 3 ^{\circ} )}{9}

2^{t a n (α)} = \frac{1}{4}

What is the general solution for 1+cos(θ)=-cos(θ) ?
The general solution for 1+cos(θ)=-cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin