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

The general solution for 3sinh(x)+cosh(x)=-2 is x=ln((-1+sqrt(3))/2)

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Popular Trigonometry >

3sinh(x)+cosh(x)=-2

Solution

3 sinh (x) + cosh (x) = - 2

Solution

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

Degrees

x = - 57.58526 \dots^{\circ}

Solution steps

3 sinh (x) + cosh (x) = - 2

Rewrite using trig identities

3 sinh (x) + cosh (x) = - 2

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

2

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

Simplify

3 (e^{x} - e^{- x}) + e^{x} + e^{- x} = - 4

Apply exponent rules

3 (e^{x} - e^{- x}) + e^{x} + e^{- x} = - 4

Apply exponent rule:

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

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

3 (e^{x} - (e^{x})^{- 1}) + e^{x} + (e^{x})^{- 1} = - 4

3 (e^{x} - (e^{x})^{- 1}) + e^{x} + (e^{x})^{- 1} = - 4

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

3 (u - \frac{1}{u}) + u + \frac{1}{u} = - 4

Multiply both sides by

u

3 (u - \frac{1}{u}) + u + \frac{1}{u} = - 4

Multiply both sides by

u

3 (u - \frac{1}{u}) u + uu + \frac{1}{u} u = - 4 u

Simplify

3 (u - \frac{1}{u}) u + uu + \frac{1}{u} u = - 4 u

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

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

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

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

Expand

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

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

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

Expand

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

3 u (u - \frac{1}{u})

Apply the distributive law:

a (b - c) = ab - a c

a = 3 u, b = u, c = \frac{1}{u}

= 3 uu - 3 u \frac{1}{u}

= 3 uu - 3 \cdot \frac{1}{u} u

Simplify

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

3 uu - 3 \cdot \frac{1}{u} u

3 uu = 3 u^{2}

3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 u^{2}

3 \cdot \frac{1}{u} u = 3

3 \cdot \frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot 3 u}{u}

Cancel the common factor:

u

= 1 \cdot 3

Multiply the numbers:

1 \cdot 3 = 3

= 3

= 3 u^{2} - 3

= 3 u^{2} - 3

= 3 u^{2} - 3 + u^{2} + 1

Simplify

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

3 u^{2} - 3 + u^{2} + 1

Group like terms

= 3 u^{2} + u^{2} - 3 + 1

Add similar elements:

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

= 4 u^{2} - 3 + 1

Add/Subtract the numbers:

- 3 + 1 = - 2

= 4 u^{2} - 2

= 4 u^{2} - 2

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

Solve

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

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

Move

4 u

to the left side

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

Add

4 u

to both sides

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

Simplify

4 u^{2} - 2 + 4 u = 0

4 u^{2} - 2 + 4 u = 0

Write in the standard form

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

4 u^{2} + 4 u - 2 = 0

Solve with the quadratic formula

4 u^{2} + 4 u - 2 = 0

Quadratic Equation Formula:

For

a = 4, b = 4, c = - 2

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

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

4^{2} - 4 \cdot 4 (- 2) = 43

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 4^{2} + 32

4^{2} = 16

= 16 + 32

Add the numbers:

16 + 32 = 48

= 48

Prime factorization of

48 : 2^{4} \cdot 3

48

48

divides by

248 = 24 \cdot 2

= 2 \cdot 24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

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

2, 3

are all prime numbers, therefore no further factorization is possible

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

= 2^{4} \cdot 3

= 2^{4} \cdot 3

Apply radical rule:

= 3 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 3

Refine

= 43

u_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 4 + 4 3}{2 \cdot 4}, u_{2} = \frac{- 4 - 4 3}{2 \cdot 4}

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

\frac{- 4 + 4 3}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 + 4 3}{8}

Factor

- 4 + 43 : 4 (- 1 + 3)

- 4 + 43

Rewrite as

= - 4 \cdot 1 + 43

Factor out common term

4

= 4 (- 1 + 3)

= \frac{4 ( - 1 + 3 )}{8}

Cancel the common factor:

4

= \frac{- 1 + 3}{2}

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

\frac{- 4 - 4 3}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 - 4 3}{8}

Factor

- 4 - 43 : - 4 (1 + 3)

- 4 - 43

Rewrite as

= - 4 \cdot 1 - 43

Factor out common term

4

= - 4 (1 + 3)

= - \frac{4 ( 1 + 3 )}{8}

Cancel the common factor:

4

= - \frac{1 + 3}{2}

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

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

Substitute back

u = e^{x},

solve for

x

Solve

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

e^{x} = \frac{- 1 + 3}{2}

Apply exponent rules

e^{x} = \frac{- 1 + 3}{2}

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

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

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

Solve

e^{x} = - \frac{1 + 3}{2} :

No Solution for

x \in R

e^{x} = - \frac{1 + 3}{2}

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3sinh(x)+cosh(x)=-2 ?
The general solution for 3sinh(x)+cosh(x)=-2 is x=ln((-1+sqrt(3))/2)