What is the general solution for 4+4cos(θ)=0 ?

The general solution for 4+4cos(θ)=0 is θ=pi+2pin

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4+4cos(θ)=0

4 + 4 cos (θ) = 0

θ = π + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 + 4 cos (θ) = 0

Move

4

to the right side

4 + 4 cos (θ) = 0

Subtract

4

from both sides

4 + 4 cos (θ) - 4 = 0 - 4

Simplify

4 cos (θ) = - 4

4 cos (θ) = - 4

Divide both sides by

4

4 cos (θ) = - 4

Divide both sides by

4

\frac{4 cos ( θ )}{4} = \frac{- 4}{4}

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

- cot (2 θ) = 0

1 - \frac{1}{2} \cdot cos (3 θ) = \frac{4 - 2}{4}, 0^{\circ} \leq θ \leq 36 0^{\circ}

\frac{cos ( x )}{sin ( x )} = - 0.00836

sin^{3} (x) = cos^{3} (x)

2 sin (θ) = 3 cos (θ) - 1, 0 <, θ < 360

What is the general solution for 4+4cos(θ)=0 ?
The general solution for 4+4cos(θ)=0 is θ=pi+2pin