What is the general solution for 1+cos(θ)=0 ?

The general solution for 1+cos(θ)=0 is θ=pi+2pin

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1+cos(θ)=0

1 + cos (θ) = 0

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos (θ) = 0

Move

1

to the right side

1 + cos (θ) = 0

Subtract

1

from both sides

1 + cos (θ) - 1 = 0 - 1

Simplify

cos (θ) = - 1

cos (θ) = - 1

General solutions for

cos (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

x, (D^{2} - 3 D + 2) y = cos (e^{- x})

2 + cos (2 x) = - 5 sin (x)

cos^{2} (θ) - \frac{1}{4} = 0

2 cot (x) + sec^{2} (x) = 0

π cos (θ) = - \frac{π}{2}

What is the general solution for 1+cos(θ)=0 ?
The general solution for 1+cos(θ)=0 is θ=pi+2pin