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

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

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0=2(1+cos(θ))

0 = 2 (1 + cos (θ))

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 2 (1 + cos (θ))

Switch sides

2 (1 + cos (θ)) = 0

Divide both sides by

2

2 (1 + cos (θ)) = 0

Divide both sides by

2

\frac{2 ( 1 + cos ( θ ) )}{2} = \frac{0}{2}

Simplify

1 + cos (θ) = 0

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

\frac{csc ( x )}{5} - 2 = 0

sin (4 y) + sin (6 y) + cos (y) = 0, 0 \leq y \leq \frac{π}{2}

2 \cdot cos^{2} (x) - sin^{2} (x) = 0

cos (θ) = - \frac{35}{6}

5 cos (x) + 2 sin (x) = 0

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