What is the general solution for cosh(npi)+sinh(npi)=0 ?

The general solution for cosh(npi)+sinh(npi)=0 is No Solution for n\in\mathbb{R}

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cosh(npi)+sinh(npi)=0

cosh (nπ) + sinh (nπ) = 0

No Solution for n \in R

Solution steps

cosh (nπ) + sinh (nπ) = 0

Rewrite using trig identities

cosh (nπ) + sinh (nπ) = 0

Use the Hyperbolic identity:

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

cosh (nπ) + \frac{e ^{nπ} - e ^{- nπ}}{2} = 0

Use the Hyperbolic identity:

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

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

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

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

No Solution for

n \in R

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

Multiply both sides by

2

\frac{e ^{nπ} + e ^{- nπ}}{2} \cdot 2 + \frac{e ^{nπ} - e ^{- nπ}}{2} \cdot 2 = 0 \cdot 2

Simplify

2 e^{nπ} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

e^{nπ} = 0

Solve

e^{nπ} = 0 :

No Solution for

n \in R

e^{nπ} = 0

a^{f (n)}

cannot be zero or negative for

n \in R

No Solution for n \in R

No Solution for n \in R

No Solution for n \in R

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3 sin (x) - cos (x) = - 1

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