What is the general solution for 2sinh(2x)-10sinh(x)=0 ?

The general solution for 2sinh(2x)-10sinh(x)=0 is x=0,x=ln((5+sqrt(21))/2),x=ln((5-sqrt(21))/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sinh(2x)-10sinh(x)=0

Solution

2 sinh (2 x) - 10 sinh (x) = 0

Solution

x = 0, x = ln (\frac{5 + 21}{2}), x = ln (\frac{5 - 21}{2})

Degrees

x = 0^{\circ}, x = 89.77098 \dots^{\circ}, x = - 89.77098 \dots^{\circ}

Solution steps

2 sinh (2 x) - 10 sinh (x) = 0

Rewrite using trig identities

2 sinh (2 x) - 10 sinh (x) = 0

Use the Hyperbolic identity:

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

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 10 \cdot \frac{e ^{x} - e ^{- x}}{2} = 0

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 10 \cdot \frac{e ^{x} - e ^{- x}}{2} = 0

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 10 \cdot \frac{e ^{x} - e ^{- x}}{2} = 0 : x = 0, x = ln (\frac{5 + 21}{2}), x = ln (\frac{5 - 21}{2})

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 10 \cdot \frac{e ^{x} - e ^{- x}}{2} = 0

Add

10 \frac{e ^{x} - e ^{- x}}{2}

to both sides

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 10 \cdot \frac{e ^{x} - e ^{- x}}{2} + 10 \cdot \frac{e ^{x} - e ^{- x}}{2} = 0 + 10 \cdot \frac{e ^{x} - e ^{- x}}{2}

Simplify

e^{2 x} - e^{- 2 x} = 5 (e^{x} - e^{- x})

Apply exponent rules

e^{2 x} - e^{- 2 x} = 5 (e^{x} - e^{- x})

Apply exponent rule:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}, e^{- x} = (e^{x})^{- 1}

(e^{x})^{2} - (e^{x})^{- 2} = 5 (e^{x} - (e^{x})^{- 1})

(e^{x})^{2} - (e^{x})^{- 2} = 5 (e^{x} - (e^{x})^{- 1})

Rewrite the equation with

e^{x} = u

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

Solve

u^{2} - u^{- 2} = 5 (u - u^{- 1}) : u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

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

Refine

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

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

5 u^{2} u

Apply exponent rule:

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

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

= 5 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 5 u^{3}

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 5 = 5

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

Cancel the common factor:

u

= 5 u

= 5 u^{3} - 5 u

= 5 u^{3} - 5 u

u^{4} - 1 = 5 u^{3} - 5 u

Solve

u^{4} - 1 = 5 u^{3} - 5 u : u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u^{4} - 1 = 5 u^{3} - 5 u

Move

5 u

to the left side

u^{4} - 1 = 5 u^{3} - 5 u

Add

5 u

to both sides

u^{4} - 1 + 5 u = 5 u^{3} - 5 u + 5 u

Simplify

u^{4} - 1 + 5 u = 5 u^{3}

u^{4} - 1 + 5 u = 5 u^{3}

Move

5 u^{3}

to the left side

u^{4} - 1 + 5 u = 5 u^{3}

Subtract

5 u^{3}

from both sides

u^{4} - 1 + 5 u - 5 u^{3} = 5 u^{3} - 5 u^{3}

Simplify

u^{4} - 1 + 5 u - 5 u^{3} = 0

u^{4} - 1 + 5 u - 5 u^{3} = 0

Write in the standard form

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

u^{4} - 5 u^{3} + 5 u - 1 = 0

Factor

u^{4} - 5 u^{3} + 5 u - 1 : (u + 1) (u - 1) (u^{2} - 5 u + 1)

u^{4} - 5 u^{3} + 5 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 1

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1}{1}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

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

\frac{u ^{4} - 5 u ^{3} + 5 u - 1}{u + 1} = u^{3} - 6 u^{2} + 6 u - 1

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

Divide

\frac{u ^{4} - 5 u ^{3} + 5 u - 1}{u + 1} : \frac{u ^{4} - 5 u ^{3} + 5 u - 1}{u + 1} = u^{3} + \frac{- 6 u ^{3} + 5 u - 1}{u + 1}

Divide the leading coefficients of the numerator

u^{4} - 5 u^{3} + 5 u - 1

and the divisor

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

Quotient = u^{3}

Multiply

u + 1

u^{3} : u^{4} + u^{3}

Subtract

u^{4} + u^{3}

from

u^{4} - 5 u^{3} + 5 u - 1

to get new remainder

Remainder = - 6 u^{3} + 5 u - 1

Therefore

\frac{u ^{4} - 5 u ^{3} + 5 u - 1}{u + 1} = u^{3} + \frac{- 6 u ^{3} + 5 u - 1}{u + 1}

= u^{3} + \frac{- 6 u ^{3} + 5 u - 1}{u + 1}

Divide

\frac{- 6 u ^{3} + 5 u - 1}{u + 1} : \frac{- 6 u ^{3} + 5 u - 1}{u + 1} = - 6 u^{2} + \frac{6 u ^{2} + 5 u - 1}{u + 1}

Divide the leading coefficients of the numerator

- 6 u^{3} + 5 u - 1

and the divisor

u + 1 : \frac{- 6 u ^{3}}{u} = - 6 u^{2}

Quotient = - 6 u^{2}

Multiply

u + 1

- 6 u^{2} : - 6 u^{3} - 6 u^{2}

Subtract

- 6 u^{3} - 6 u^{2}

from

- 6 u^{3} + 5 u - 1

to get new remainder

Remainder = 6 u^{2} + 5 u - 1

Therefore

\frac{- 6 u ^{3} + 5 u - 1}{u + 1} = - 6 u^{2} + \frac{6 u ^{2} + 5 u - 1}{u + 1}

= u^{3} - 6 u^{2} + \frac{6 u ^{2} + 5 u - 1}{u + 1}

Divide

\frac{6 u ^{2} + 5 u - 1}{u + 1} : \frac{6 u ^{2} + 5 u - 1}{u + 1} = 6 u + \frac{- u - 1}{u + 1}

Divide the leading coefficients of the numerator

6 u^{2} + 5 u - 1

and the divisor

u + 1 : \frac{6 u ^{2}}{u} = 6 u

Quotient = 6 u

Multiply

u + 1

6 u : 6 u^{2} + 6 u

Subtract

6 u^{2} + 6 u

from

6 u^{2} + 5 u - 1

to get new remainder

Remainder = - u - 1

Therefore

\frac{6 u ^{2} + 5 u - 1}{u + 1} = 6 u + \frac{- u - 1}{u + 1}

= u^{3} - 6 u^{2} + 6 u + \frac{- u - 1}{u + 1}

Divide

\frac{- u - 1}{u + 1} : \frac{- u - 1}{u + 1} = - 1

Divide the leading coefficients of the numerator

- u - 1

and the divisor

u + 1 : \frac{- u}{u} = - 1

Quotient = - 1

Multiply

u + 1

- 1 : - u - 1

Subtract

- u - 1

from

- u - 1

to get new remainder

Remainder = 0

Therefore

\frac{- u - 1}{u + 1} = - 1

= u^{3} - 6 u^{2} + 6 u - 1

= u^{3} - 6 u^{2} + 6 u - 1

Factor

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

u^{3} - 6 u^{2} + 6 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 1

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1}{1}

\frac{1}{1}

is a root of the expression, so factor out

u - 1

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

\frac{u ^{3} - 6 u ^{2} + 6 u - 1}{u - 1} = u^{2} - 5 u + 1

\frac{u ^{3} - 6 u ^{2} + 6 u - 1}{u - 1}

Divide

\frac{u ^{3} - 6 u ^{2} + 6 u - 1}{u - 1} : \frac{u ^{3} - 6 u ^{2} + 6 u - 1}{u - 1} = u^{2} + \frac{- 5 u ^{2} + 6 u - 1}{u - 1}

Divide the leading coefficients of the numerator

u^{3} - 6 u^{2} + 6 u - 1

and the divisor

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

Quotient = u^{2}

Multiply

u - 1

u^{2} : u^{3} - u^{2}

Subtract

u^{3} - u^{2}

from

u^{3} - 6 u^{2} + 6 u - 1

to get new remainder

Remainder = - 5 u^{2} + 6 u - 1

Therefore

\frac{u ^{3} - 6 u ^{2} + 6 u - 1}{u - 1} = u^{2} + \frac{- 5 u ^{2} + 6 u - 1}{u - 1}

= u^{2} + \frac{- 5 u ^{2} + 6 u - 1}{u - 1}

Divide

\frac{- 5 u ^{2} + 6 u - 1}{u - 1} : \frac{- 5 u ^{2} + 6 u - 1}{u - 1} = - 5 u + \frac{u - 1}{u - 1}

Divide the leading coefficients of the numerator

- 5 u^{2} + 6 u - 1

and the divisor

u - 1 : \frac{- 5 u ^{2}}{u} = - 5 u

Quotient = - 5 u

Multiply

u - 1

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

Subtract

- 5 u^{2} + 5 u

from

- 5 u^{2} + 6 u - 1

to get new remainder

Remainder = u - 1

Therefore

\frac{- 5 u ^{2} + 6 u - 1}{u - 1} = - 5 u + \frac{u - 1}{u - 1}

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

Divide

\frac{u - 1}{u - 1} : \frac{u - 1}{u - 1} = 1

Divide the leading coefficients of the numerator

u - 1

and the divisor

u - 1 : \frac{u}{u} = 1

Quotient = 1

Multiply

u - 1

1 : u - 1

Subtract

u - 1

from

u - 1

to get new remainder

Remainder = 0

Therefore

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

= u^{2} - 5 u + 1

= u^{2} - 5 u + 1

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

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

(u + 1) (u - 1) (u^{2} - 5 u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or u - 1 = 0 or u^{2} - 5 u + 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

u^{2} - 5 u + 1 = 0 : u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u^{2} - 5 u + 1 = 0

Solve with the quadratic formula

u^{2} - 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 5, c = 1

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

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

(- 5)^{2} - 4 \cdot 1 \cdot 1 = 21

(- 5)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 5^{2} - 4

5^{2} = 25

= 25 - 4

Subtract the numbers:

25 - 4 = 21

= 21

u_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

u = \frac{- ( - 5 ) + 21}{2 \cdot 1} : \frac{5 + 21}{2}

\frac{- ( - 5 ) + 21}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 + 21}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 + 21}{2}

u = \frac{- ( - 5 ) - 21}{2 \cdot 1} : \frac{5 - 21}{2}

\frac{- ( - 5 ) - 21}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 - 21}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 - 21}{2}

The solutions to the quadratic equation are:

u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

The solutions are

u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

5 (u - u^{- 1})

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u = - 1, u = 1, u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

Substitute back

u = e^{x},

solve for

x

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

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} = \frac{5 + 21}{2} : x = ln (\frac{5 + 21}{2})

e^{x} = \frac{5 + 21}{2}

Apply exponent rules

e^{x} = \frac{5 + 21}{2}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5 + 21}{2})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{5 + 21}{2})

x = ln (\frac{5 + 21}{2})

Solve

e^{x} = \frac{5 - 21}{2} : x = ln (\frac{5 - 21}{2})

e^{x} = \frac{5 - 21}{2}

Apply exponent rules

e^{x} = \frac{5 - 21}{2}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5 - 21}{2})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{5 - 21}{2})

x = ln (\frac{5 - 21}{2})

x = 0, x = ln (\frac{5 + 21}{2}), x = ln (\frac{5 - 21}{2})

x = 0, x = ln (\frac{5 + 21}{2}), x = ln (\frac{5 - 21}{2})

Graph

Popular Examples

cot(x-pi/2)+1=0

cot (x - \frac{π}{2}) + 1 = 0

-sin(t)+cos(t)=0

- sin (t) + cos (t) = 0

4cos(x)=3sec(x)

4 cos (x) = 3 sec (x)

2sin(x)cos(x)=2sin(x)

2 sin (x) cos (x) = 2 sin (x)

sin^2(θ)=3-2cos^2(θ)

sin^{2} (θ) = 3 - 2 cos^{2} (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 2sinh(2x)-10sinh(x)=0 ?
The general solution for 2sinh(2x)-10sinh(x)=0 is x=0,x=ln((5+sqrt(21))/2),x=ln((5-sqrt(21))/2)