What is the general solution for 4cos(z)+5=0 ?

The general solution for 4cos(z)+5=0 is No Solution for z\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos(z)+5=0

4 cos (z) + 5 = 0

No Solution for z \in R

Solution steps

4 cos (z) + 5 = 0

Move

5

to the right side

4 cos (z) + 5 = 0

Subtract

5

from both sides

4 cos (z) + 5 - 5 = 0 - 5

Simplify

4 cos (z) = - 5

4 cos (z) = - 5

Divide both sides by

4

4 cos (z) = - 5

Divide both sides by

4

\frac{4 cos ( z )}{4} = \frac{- 5}{4}

Simplify

cos (z) = - \frac{5}{4}

cos (z) = - \frac{5}{4}

- 1 \leq cos (x) \leq 1

No Solution for z \in R

y, sin (x^{2} y^{2}) = x

5 + 4 cos (θ) = 0

- sec^{2} (x) + tan (x) + 1 = 0

tan (x) = - 0.8

cos^{3} (x) - 1 = 0

What is the general solution for 4cos(z)+5=0 ?
The general solution for 4cos(z)+5=0 is No Solution for z\in\mathbb{R}