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

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

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cosh(z)=0

cosh (z) = 0

No Solution for z \in R

Solution steps

cosh (z) = 0

Rewrite using trig identities

cosh (z) = 0

Use the Hyperbolic identity:

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

\frac{e ^{z} + e ^{- z}}{2} = 0

\frac{e ^{z} + e ^{- z}}{2} = 0

\frac{e ^{z} + e ^{- z}}{2} = 0 : z =

Undefined

\frac{e ^{z} + e ^{- z}}{2} = 0

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

e^{z} + e^{- z} = 0

Subtract

e^{- z}

from both sides

e^{z} + e^{- z} - e^{- z} = 0 - e^{- z}

Simplify

e^{z} = - e^{- z}

Apply exponent rules

e^{z} = - e^{- z}

f (x) = g (x)

, then

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

ln (e^{z}) = ln (- e^{- z})

Apply log rule:

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

ln (- e^{- z}) = ln (- 1) + ln (e^{- z})

ln (e^{z}) = ln (- 1) + ln (e^{- z})

Apply log rule:

ln (e^{a}) = a

ln (e^{z}) = z

z = ln (- 1) + ln (e^{- z})

Apply log rule:

ln (e^{a}) = a

ln (e^{- z}) = - z

z = ln (- 1) - z

z = ln (- 1) - z

Solve

z = ln (- 1) - z : z =

Undefined

z = ln (- 1) - z

Simplify

ln (- 1) - z :

Undefined

ln (- 1) - z

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

is undefined

= Undefined

z = Undefined

z = Undefined

z = Undefined

Since the equation is undefined for:

Undefined

No Solution for z \in R

cos (x) = \frac{2}{π}

tan^{2} (x) + 3 tan (x) + 1 = 0

csc (2 x) = 2, 0 \leq x < 2 π

sin (x) + cos (x) = 1.2

30 cos (30 x) + 14 = - 16

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