What is the general solution for cosh(4x)=0 ?

The general solution for cosh(4x)=0 is No Solution for x\in\mathbb{R}

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cosh(4x)=0

cosh (4 x) = 0

No Solution for x \in R

Solution steps

cosh (4 x) = 0

Rewrite using trig identities

cosh (4 x) = 0

Use the Hyperbolic identity:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 0

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 0

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 0 : x = \frac{Undefined}{4}

\frac{e ^{4 x} + e ^{- 4 x}}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

e^{4 x} + e^{- 4 x} = 0

Subtract

e^{- 4 x}

from both sides

e^{4 x} + e^{- 4 x} - e^{- 4 x} = 0 - e^{- 4 x}

Simplify

e^{4 x} = - e^{- 4 x}

Apply exponent rules

e^{4 x} = - e^{- 4 x}

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{4 x}) = ln (- e^{- 4 x})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (- e^{- 4 x}) = ln (- 1) + ln (e^{- 4 x})

ln (e^{4 x}) = ln (- 1) + ln (e^{- 4 x})

Apply log rule:

ln (e^{a}) = a

ln (e^{4 x}) = 4 x

4 x = ln (- 1) + ln (e^{- 4 x})

Apply log rule:

ln (e^{a}) = a

ln (e^{- 4 x}) = - 4 x

4 x = ln (- 1) - 4 x

4 x = ln (- 1) - 4 x

Solve

4 x = ln (- 1) - 4 x : x = \frac{Undefined}{4}

4 x = ln (- 1) - 4 x

Simplify

ln (- 1) - 4 x :

Undefined

ln (- 1) - 4 x

lo g_{a} (b), b < 0

is undefined

= Undefined

4 x = Undefined

Divide both sides by

4

4 x = Undefined

Divide both sides by

4

\frac{4 x}{4} = \frac{Undefined}{4}

Simplify

x = \frac{Undefined}{4}

x = \frac{Undefined}{4}

x = \frac{Undefined}{4}

x = \frac{Undefined}{4}

Since the equation is undefined for:

\frac{Undefined}{4}

No Solution for x \in R

x, tan (2 x) = - \frac{( \frac{30 + 20}{2} )}{80}

4 cos (3 θ) = 2

sin (x) = - 0.8176

sin (x) = - 0.342

(cos (x) - 4) (cos (x) + 1) = 0

What is the general solution for cosh(4x)=0 ?
The general solution for cosh(4x)=0 is No Solution for x\in\mathbb{R}