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

The general solution for 4+2cos(θ)=0 is No Solution for θ\in\mathbb{R}

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4+2cos(θ)=0

4 + 2 cos (θ) = 0

No Solution for θ \in R

Solution steps

4 + 2 cos (θ) = 0

Move

4

to the right side

4 + 2 cos (θ) = 0

Subtract

4

from both sides

4 + 2 cos (θ) - 4 = 0 - 4

Simplify

2 cos (θ) = - 4

2 cos (θ) = - 4

Divide both sides by

2

2 cos (θ) = - 4

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 4}{2}

Simplify

\frac{2 cos ( θ )}{2} = \frac{- 4}{2}

Simplify

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos (θ)

Simplify

\frac{- 4}{2} : - 2

\frac{- 4}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

cos (θ) = - 2

cos (θ) = - 2

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

9.81 sin (θ) - 5.88 cos (θ) = 2.77

cos (θ) = π

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

sin (x) - cos (x) = 1, 0 \leq x \leq 2 π

0 = sin (0 + c)

What is the general solution for 4+2cos(θ)=0 ?
The general solution for 4+2cos(θ)=0 is No Solution for θ\in\mathbb{R}