What is the general solution for 4cosh(2x)=4+sinh(2x) ?

The general solution for 4cosh(2x)=4+sinh(2x) is x= 1/2 ln(5/3),x=0

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4cosh(2x)=4+sinh(2x)

Solution

4 cosh (2 x) = 4 + sinh (2 x)

Solution

x = \frac{1}{2} ln (\frac{5}{3}), x = 0

Degrees

x = 14.63407 \dots^{\circ}, x = 0^{\circ}

Solution steps

4 cosh (2 x) = 4 + sinh (2 x)

Rewrite using trig identities

4 cosh (2 x) = 4 + sinh (2 x)

Use the Hyperbolic identity:

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

4 cosh (2 x) = 4 + \frac{e ^{2 x} - e ^{- 2 x}}{2}

Use the Hyperbolic identity:

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

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

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

4 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4 + \frac{e ^{2 x} - e ^{- 2 x}}{2} : x = \frac{1}{2} ln (\frac{5}{3}), x = 0

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

Multiply both sides by

2

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

Simplify

4 (e^{2 x} + e^{- 2 x}) = 8 + e^{2 x} - e^{- 2 x}

Apply exponent rules

4 (e^{2 x} + e^{- 2 x}) = 8 + e^{2 x} - e^{- 2 x}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}

4 ((e^{x})^{2} + (e^{x})^{- 2}) = 8 + (e^{x})^{2} - (e^{x})^{- 2}

4 ((e^{x})^{2} + (e^{x})^{- 2}) = 8 + (e^{x})^{2} - (e^{x})^{- 2}

Rewrite the equation with

e^{x} = u

4 ((u)^{2} + (u)^{- 2}) = 8 + (u)^{2} - (u)^{- 2}

Solve

4 (u^{2} + u^{- 2}) = 8 + u^{2} - u^{- 2} : u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

4 (u^{2} + u^{- 2}) = 8 + u^{2} - u^{- 2}

Refine

4 (u^{2} + \frac{1}{u ^{2}}) = 8 + u^{2} - \frac{1}{u ^{2}}

Multiply both sides by

u^{2}

4 (u^{2} + \frac{1}{u ^{2}}) = 8 + u^{2} - \frac{1}{u ^{2}}

Multiply both sides by

u^{2}

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} = 8 u^{2} + u^{2} u^{2} - \frac{1}{u ^{2}} u^{2}

Simplify

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} = 8 u^{2} + u^{2} u^{2} - \frac{1}{u ^{2}} u^{2}

Simplify

u^{2} u^{2} : u^{4}

u^{2} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

Simplify

- \frac{1}{u ^{2}} u^{2} : - 1

- \frac{1}{u ^{2}} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u ^{2}}{u ^{2}}

Cancel the common factor:

u^{2}

= - 1

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} = 8 u^{2} + u^{4} - 1

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} = 8 u^{2} + u^{4} - 1

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} = 8 u^{2} + u^{4} - 1

Expand

4 (u^{2} + \frac{1}{u ^{2}}) u^{2} : 4 u^{4} + 4

4 (u^{2} + \frac{1}{u ^{2}}) u^{2}

= 4 u^{2} (u^{2} + \frac{1}{u ^{2}})

Apply the distributive law:

a (b + c) = ab + a c

a = 4 u^{2}, b = u^{2}, c = \frac{1}{u ^{2}}

= 4 u^{2} u^{2} + 4 u^{2} \frac{1}{u ^{2}}

= 4 u^{2} u^{2} + 4 \cdot \frac{1}{u ^{2}} u^{2}

Simplify

4 u^{2} u^{2} + 4 \cdot \frac{1}{u ^{2}} u^{2} : 4 u^{4} + 4

4 u^{2} u^{2} + 4 \cdot \frac{1}{u ^{2}} u^{2}

4 u^{2} u^{2} = 4 u^{4}

4 u^{2} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= 4 u^{2 + 2}

Add the numbers:

2 + 2 = 4

= 4 u^{4}

4 \cdot \frac{1}{u ^{2}} u^{2} = 4

4 \cdot \frac{1}{u ^{2}} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 u ^{2}}{u ^{2}}

Cancel the common factor:

u^{2}

= 1 \cdot 4

Multiply the numbers:

1 \cdot 4 = 4

= 4

= 4 u^{4} + 4

= 4 u^{4} + 4

4 u^{4} + 4 = 8 u^{2} + u^{4} - 1

Move

1

to the left side

4 u^{4} + 4 = 8 u^{2} + u^{4} - 1

Add

1

to both sides

4 u^{4} + 4 + 1 = 8 u^{2} + u^{4} - 1 + 1

Simplify

4 u^{4} + 5 = 8 u^{2} + u^{4}

4 u^{4} + 5 = 8 u^{2} + u^{4}

Solve

4 u^{4} + 5 = 8 u^{2} + u^{4} : u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

4 u^{4} + 5 = 8 u^{2} + u^{4}

Move

u^{4}

to the left side

4 u^{4} + 5 = 8 u^{2} + u^{4}

Subtract

u^{4}

from both sides

4 u^{4} + 5 - u^{4} = 8 u^{2} + u^{4} - u^{4}

Simplify

3 u^{4} + 5 = 8 u^{2}

3 u^{4} + 5 = 8 u^{2}

Move

8 u^{2}

to the left side

3 u^{4} + 5 = 8 u^{2}

Subtract

8 u^{2}

from both sides

3 u^{4} + 5 - 8 u^{2} = 8 u^{2} - 8 u^{2}

Simplify

3 u^{4} + 5 - 8 u^{2} = 0

3 u^{4} + 5 - 8 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

3 u^{4} - 8 u^{2} + 5 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

3 v^{2} - 8 v + 5 = 0

Solve

3 v^{2} - 8 v + 5 = 0 : v = \frac{5}{3}, v = 1

3 v^{2} - 8 v + 5 = 0

Solve with the quadratic formula

3 v^{2} - 8 v + 5 = 0

Quadratic Equation Formula:

For

a = 3, b = - 8, c = 5

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 \cdot 5 = 2

(- 8)^{2} - 4 \cdot 3 \cdot 5

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 3 \cdot 5

Multiply the numbers:

4 \cdot 3 \cdot 5 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

Subtract the numbers:

64 - 60 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

v_{1, 2} = \frac{- ( - 8 ) \pm 2}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- ( - 8 ) + 2}{2 \cdot 3}, v_{2} = \frac{- ( - 8 ) - 2}{2 \cdot 3}

v = \frac{- ( - 8 ) + 2}{2 \cdot 3} : \frac{5}{3}

\frac{- ( - 8 ) + 2}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 + 2}{2 \cdot 3}

Add the numbers:

8 + 2 = 10

= \frac{10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{10}{6}

Cancel the common factor:

2

= \frac{5}{3}

v = \frac{- ( - 8 ) - 2}{2 \cdot 3} : 1

\frac{- ( - 8 ) - 2}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 - 2}{2 \cdot 3}

Subtract the numbers:

8 - 2 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

v = \frac{5}{3}, v = 1

v = \frac{5}{3}, v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{5}{3} : u = \frac{5}{3}, u = - \frac{5}{3}

u^{2} = \frac{5}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{5}{3}, u = - \frac{5}{3}

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

4 (u^{2} + u^{- 2})

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Take the denominator(s) of

8 + u^{2} - u^{- 2}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

u = \frac{5}{3}, u = - \frac{5}{3}, u = 1, u = - 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{5}{3} : x = \frac{1}{2} ln (\frac{5}{3})

e^{x} = \frac{5}{3}

Apply exponent rules

e^{x} = \frac{5}{3}

Apply exponent rule:

a = a^{\frac{1}{2}}

\frac{5}{3} = (\frac{5}{3})^{\frac{1}{2}}

e^{x} = (\frac{5}{3})^{\frac{1}{2}}

f (x) = g (x)

, then

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

ln (e^{x}) = ln ((\frac{5}{3})^{\frac{1}{2}})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln ((\frac{5}{3})^{\frac{1}{2}})

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

ln ((\frac{5}{3})^{\frac{1}{2}}) = \frac{1}{2} ln (\frac{5}{3})

x = \frac{1}{2} ln (\frac{5}{3})

x = \frac{1}{2} ln (\frac{5}{3})

Solve

e^{x} = - \frac{5}{3} :

No Solution for

x \in R

e^{x} = - \frac{5}{3}

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

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

ln (e^{x}) = ln (1)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

Solve

e^{x} = - 1 :

No Solution for

x \in R

e^{x} = - 1

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = \frac{1}{2} ln (\frac{5}{3}), x = 0

x = \frac{1}{2} ln (\frac{5}{3}), x = 0

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Frequently Asked Questions (FAQ)

What is the general solution for 4cosh(2x)=4+sinh(2x) ?
The general solution for 4cosh(2x)=4+sinh(2x) is x= 1/2 ln(5/3),x=0