What is the general solution for 0=2+2cos(θ) ?

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

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0=2+2cos(θ)

0 = 2 + 2 cos (θ)

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 2 + 2 cos (θ)

Switch sides

2 + 2 cos (θ) = 0

Move

2

to the right side

2 + 2 cos (θ) = 0

Subtract

2

from both sides

2 + 2 cos (θ) - 2 = 0 - 2

Simplify

2 cos (θ) = - 2

2 cos (θ) = - 2

Divide both sides by

2

2 cos (θ) = - 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 2}{2}

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

cos (4 5^{\circ} + x) - cos (4 5^{\circ} - x) = 2 cos (x)

sin (b) = cos (2 b - 30)

2 sin (x) cos (x) - 4 cos (x) - sin (x) + 2 = 0

\frac{sin ( \frac{π}{2} )}{5} = \frac{sin ( x )}{4}

cos (3 x - 1) = \frac{1}{2}

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