What is the general solution for 2cos(θ)+5=0 ?

The general solution for 2cos(θ)+5=0 is No Solution for θ\in\mathbb{R}

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2cos(θ)+5=0

2 cos (θ) + 5 = 0

No Solution for θ \in R

Solution steps

2 cos (θ) + 5 = 0

Move

5

to the right side

2 cos (θ) + 5 = 0

Subtract

5

from both sides

2 cos (θ) + 5 - 5 = 0 - 5

Simplify

2 cos (θ) = - 5

2 cos (θ) = - 5

Divide both sides by

2

2 cos (θ) = - 5

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 5}{2}

Simplify

cos (θ) = - \frac{5}{2}

cos (θ) = - \frac{5}{2}

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

sin (x) cos (x) = cos (x)

2 cos (2 θ) + 1 = 0, 0 \leq θ \leq 2 π

cot (x) sin (x) + cot (x) = 0

4 cos (2 θ) = 4 sin (2 θ)

tan (x) = - 3, 0 \leq x \leq 2 π

What is the general solution for 2cos(θ)+5=0 ?
The general solution for 2cos(θ)+5=0 is No Solution for θ\in\mathbb{R}