What is the general solution for 4cosh(x)+3sinh(x)=5 ?

The general solution for 4cosh(x)+3sinh(x)=5 is x=ln((5+3sqrt(2))/7),x=ln((5-3sqrt(2))/7)

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Popular Trigonometry >

4cosh(x)+3sinh(x)=5

Solution

4 cosh (x) + 3 sinh (x) = 5

Solution

x = ln (\frac{5 + 3 2}{7}), x = ln (\frac{5 - 3 2}{7})

Degrees

x = 15.92349 \dots^{\circ}, x = - 127.41593 \dots^{\circ}

Solution steps

4 cosh (x) + 3 sinh (x) = 5

Rewrite using trig identities

4 cosh (x) + 3 sinh (x) = 5

Use the Hyperbolic identity:

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

4 cosh (x) + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Use the Hyperbolic identity:

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

4 \cdot \frac{e ^{x} + e ^{- x}}{2} + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

4 \cdot \frac{e ^{x} + e ^{- x}}{2} + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

4 \cdot \frac{e ^{x} + e ^{- x}}{2} + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5 : x = ln (\frac{5 + 3 2}{7}), x = ln (\frac{5 - 3 2}{7})

4 \cdot \frac{e ^{x} + e ^{- x}}{2} + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Apply exponent rules

4 \cdot \frac{e ^{x} + e ^{- x}}{2} + 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

4 \cdot \frac{u + u ^{- 1}}{2} + 3 \cdot \frac{u - u ^{- 1}}{2} = 5 : u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

4 \cdot \frac{u + u ^{- 1}}{2} + 3 \cdot \frac{u - u ^{- 1}}{2} = 5

Refine

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

Multiply by LCM

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

Find Least Common Multiplier of

u, 2 u : 2 u

u, 2 u

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

u

2 u

= 2 u

Multiply by LCM=

2 u

\frac{2 ( u ^{2} + 1 )}{u} \cdot 2 u + \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u = 5 \cdot 2 u

Simplify

\frac{2 ( u ^{2} + 1 )}{u} \cdot 2 u + \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u = 5 \cdot 2 u

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

u

= 2 (u^{2} + 1) \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4 (u^{2} + 1)

Simplify

\frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u : 3 (u^{2} - 1)

\frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u

Multiply fractions:

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

= \frac{3 ( u ^{2} - 1 ) \cdot 2 u}{2 u}

Cancel the common factor:

2

= \frac{3 ( u ^{2} - 1 ) u}{u}

Cancel the common factor:

u

= 3 (u^{2} - 1)

Simplify

5 \cdot 2 u : 10 u

5 \cdot 2 u

Multiply the numbers:

5 \cdot 2 = 10

= 10 u

4 (u^{2} + 1) + 3 (u^{2} - 1) = 10 u

4 (u^{2} + 1) + 3 (u^{2} - 1) = 10 u

4 (u^{2} + 1) + 3 (u^{2} - 1) = 10 u

Solve

4 (u^{2} + 1) + 3 (u^{2} - 1) = 10 u : u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

4 (u^{2} + 1) + 3 (u^{2} - 1) = 10 u

Expand

4 (u^{2} + 1) + 3 (u^{2} - 1) : 7 u^{2} + 1

4 (u^{2} + 1) + 3 (u^{2} - 1)

Expand

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

4 (u^{2} + 1)

Apply the distributive law:

a (b + c) = ab + a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

= 4 u^{2} + 4

= 4 u^{2} + 4 + 3 (u^{2} - 1)

Expand

3 (u^{2} - 1) : 3 u^{2} - 3

3 (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = u^{2}, c = 1

= 3 u^{2} - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 u^{2} - 3

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

Simplify

4 u^{2} + 4 + 3 u^{2} - 3 : 7 u^{2} + 1

4 u^{2} + 4 + 3 u^{2} - 3

Group like terms

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

Add similar elements:

4 u^{2} + 3 u^{2} = 7 u^{2}

= 7 u^{2} + 4 - 3

Add/Subtract the numbers:

4 - 3 = 1

= 7 u^{2} + 1

= 7 u^{2} + 1

7 u^{2} + 1 = 10 u

Move

10 u

to the left side

7 u^{2} + 1 = 10 u

Subtract

10 u

from both sides

7 u^{2} + 1 - 10 u = 10 u - 10 u

Simplify

7 u^{2} + 1 - 10 u = 0

7 u^{2} + 1 - 10 u = 0

Write in the standard form

a x^{2} + b x + c = 0

7 u^{2} - 10 u + 1 = 0

Solve with the quadratic formula

7 u^{2} - 10 u + 1 = 0

Quadratic Equation Formula:

For

a = 7, b = - 10, c = 1

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

(- 10)^{2} - 4 \cdot 7 \cdot 1 = 62

(- 10)^{2} - 4 \cdot 7 \cdot 1

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 7 \cdot 1

Multiply the numbers:

4 \cdot 7 \cdot 1 = 28

= 1 0^{2} - 28

1 0^{2} = 100

= 100 - 28

Subtract the numbers:

100 - 28 = 72

= 72

Prime factorization of

72 : 2^{3} \cdot 3^{2}

72

72

divides by

272 = 36 \cdot 2

= 2 \cdot 36

36

divides by

236 = 18 \cdot 2

= 2 \cdot 2 \cdot 18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{2}

= 2^{3} \cdot 3^{2}

Apply exponent rule:

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

= 2^{2} \cdot 3^{2} \cdot 2

Apply radical rule:

= 2 2^{2} 3^{2}

Apply radical rule:

2^{2} = 2

= 22 3^{2}

Apply radical rule:

3^{2} = 3

= 2 \cdot 32

Refine

= 62

u_{1, 2} = \frac{- ( - 10 ) \pm 6 2}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 6 2}{2 \cdot 7}, u_{2} = \frac{- ( - 10 ) - 6 2}{2 \cdot 7}

u = \frac{- ( - 10 ) + 6 2}{2 \cdot 7} : \frac{5 + 3 2}{7}

\frac{- ( - 10 ) + 6 2}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{10 + 6 2}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{10 + 6 2}{14}

Factor

10 + 62 : 2 (5 + 32)

10 + 62

Rewrite as

= 2 \cdot 5 + 2 \cdot 32

Factor out common term

2

= 2 (5 + 32)

= \frac{2 ( 5 + 3 2 )}{14}

Cancel the common factor:

2

= \frac{5 + 3 2}{7}

u = \frac{- ( - 10 ) - 6 2}{2 \cdot 7} : \frac{5 - 3 2}{7}

\frac{- ( - 10 ) - 6 2}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{10 - 6 2}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{10 - 6 2}{14}

Factor

10 - 62 : 2 (5 - 32)

10 - 62

Rewrite as

= 2 \cdot 5 - 2 \cdot 32

Factor out common term

2

= 2 (5 - 32)

= \frac{2 ( 5 - 3 2 )}{14}

Cancel the common factor:

2

= \frac{5 - 3 2}{7}

The solutions to the quadratic equation are:

u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

4 \frac{u + u ^{- 1}}{2} + 3 \frac{u - u ^{- 1}}{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

u = \frac{5 + 3 2}{7}, u = \frac{5 - 3 2}{7}

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{5 + 3 2}{7} : x = ln (\frac{5 + 3 2}{7})

e^{x} = \frac{5 + 3 2}{7}

Apply exponent rules

e^{x} = \frac{5 + 3 2}{7}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5 + 3 2}{7})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{5 + 3 2}{7})

x = ln (\frac{5 + 3 2}{7})

Solve

e^{x} = \frac{5 - 3 2}{7} : x = ln (\frac{5 - 3 2}{7})

e^{x} = \frac{5 - 3 2}{7}

Apply exponent rules

e^{x} = \frac{5 - 3 2}{7}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5 - 3 2}{7})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{5 - 3 2}{7})

x = ln (\frac{5 - 3 2}{7})

x = ln (\frac{5 + 3 2}{7}), x = ln (\frac{5 - 3 2}{7})

x = ln (\frac{5 + 3 2}{7}), x = ln (\frac{5 - 3 2}{7})

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cosh(x)+3sinh(x)=5 ?
The general solution for 4cosh(x)+3sinh(x)=5 is x=ln((5+3sqrt(2))/7),x=ln((5-3sqrt(2))/7)