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

The general solution for 3cosh(x)+5sinh(x)=-3 is x=-2ln(2)

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Popular Trigonometry >

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

Solution

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

Solution

x = - 2 ln (2)

Degrees

x = - 79.42881 \dots^{\circ}

Solution steps

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

Rewrite using trig identities

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

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

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

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Multiply both sides by

2 u

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

Multiply both sides by

2 u

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

Simplify

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

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

\frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u : 5 (u^{2} - 1)

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

u

= 5 (u^{2} - 1)

Simplify

- 3 \cdot 2 u : - 6 u

- 3 \cdot 2 u

Multiply the numbers:

3 \cdot 2 = 6

= - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

Solve

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u : u = \frac{1}{4}, u = - 1

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

Expand

3 (u^{2} + 1) + 5 (u^{2} - 1) : 8 u^{2} - 2

3 (u^{2} + 1) + 5 (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

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

Expand

5 (u^{2} - 1) : 5 u^{2} - 5

5 (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = u^{2}, c = 1

= 5 u^{2} - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 u^{2} - 5

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

Simplify

3 u^{2} + 3 + 5 u^{2} - 5 : 8 u^{2} - 2

3 u^{2} + 3 + 5 u^{2} - 5

Group like terms

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

Add similar elements:

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

= 8 u^{2} + 3 - 5

Add/Subtract the numbers:

3 - 5 = - 2

= 8 u^{2} - 2

= 8 u^{2} - 2

8 u^{2} - 2 = - 6 u

Move

6 u

to the left side

8 u^{2} - 2 = - 6 u

Add

6 u

to both sides

8 u^{2} - 2 + 6 u = - 6 u + 6 u

Simplify

8 u^{2} - 2 + 6 u = 0

8 u^{2} - 2 + 6 u = 0

Write in the standard form

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

8 u^{2} + 6 u - 2 = 0

Solve with the quadratic formula

8 u^{2} + 6 u - 2 = 0

Quadratic Equation Formula:

For

a = 8, b = 6, c = - 2

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

6^{2} - 4 \cdot 8 (- 2) = 10

6^{2} - 4 \cdot 8 (- 2)

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 8 \cdot 2

Multiply the numbers:

4 \cdot 8 \cdot 2 = 64

= 6^{2} + 64

6^{2} = 36

= 36 + 64

Add the numbers:

36 + 64 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 6 \pm 10}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 6 + 10}{2 \cdot 8}, u_{2} = \frac{- 6 - 10}{2 \cdot 8}

u = \frac{- 6 + 10}{2 \cdot 8} : \frac{1}{4}

\frac{- 6 + 10}{2 \cdot 8}

Add/Subtract the numbers:

- 6 + 10 = 4

= \frac{4}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{4}{16}

Cancel the common factor:

4

= \frac{1}{4}

u = \frac{- 6 - 10}{2 \cdot 8} : - 1

\frac{- 6 - 10}{2 \cdot 8}

Subtract the numbers:

- 6 - 10 = - 16

= \frac{- 16}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 16}{16}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{4}, u = - 1

u = \frac{1}{4}, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

u = \frac{1}{4}, u = - 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{1}{4} : x = - 2 ln (2)

e^{x} = \frac{1}{4}

Apply exponent rules

e^{x} = \frac{1}{4}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{1}{4})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{1}{4})

Simplify

ln (\frac{1}{4}) : - 2 ln (2)

ln (\frac{1}{4})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (4)

Rewrite

4

in power-base form:

4 = 2^{2}

= - ln (2^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (2^{2}) = 2 ln (2)

= - 2 ln (2)

x = - 2 ln (2)

x = - 2 ln (2)

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 = - 2 ln (2)

x = - 2 ln (2)

Graph

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sin(u)= 5/13

Frequently Asked Questions (FAQ)

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