What is the general solution for 1-sinh^2(x)=0 ?

The general solution for 1-sinh^2(x)=0 is x=ln(-1+sqrt(2)),x=ln(1+sqrt(2))

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Popular >

1-sinh^2(x)=0

Solution

1 - sinh^{2} (x) = 0

Solution

x = ln (- 1 + 2), x = ln (1 + 2)

Degrees

x = - 50.49898 \dots^{\circ}, x = 50.49898 \dots^{\circ}

Solution steps

1 - sinh^{2} (x) = 0

Rewrite using trig identities

1 - sinh^{2} (x) = 0

Use the Hyperbolic identity:

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

1 - (\frac{e ^{x} - e ^{- x}}{2})^{2} = 0

1 - (\frac{e ^{x} - e ^{- x}}{2})^{2} = 0

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

1 - (\frac{e ^{x} - e ^{- x}}{2})^{2} = 0

Factor

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

1 - (\frac{e ^{x} - e ^{- x}}{2})^{2}

Factor out common term

- 1

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

Factor

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

(\frac{e ^{x} - e ^{- x}}{2})^{2} - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Refine

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

Factor

e^{x} - e^{- x} : e^{- x} (e^{x} + 1) (e^{x} - 1)

e^{x} - e^{- x}

Apply exponent rule:

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

e^{x} = e^{- x} e^{2 x}

= e^{- x} e^{2 x} - e^{- x}

Factor out common term

e^{- x}

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

Factor

e^{2 x} - 1 : (e^{x} + 1) (e^{x} - 1)

e^{2 x} - 1

Apply exponent rule:

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

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

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

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

Factor

e^{x} - e^{- x} : e^{- x} (e^{x} + 1) (e^{x} - 1)

e^{x} - e^{- x}

Apply exponent rule:

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

e^{x} = e^{- x} e^{2 x}

= e^{- x} e^{2 x} - e^{- x}

Factor out common term

e^{- x}

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

Factor

e^{2 x} - 1 : (e^{x} + 1) (e^{x} - 1)

e^{2 x} - 1

Apply exponent rule:

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

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

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

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

- (\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} + 1) (\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} + 1 = 0 or \frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} - 1 = 0

Solve

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} + 1 = 0 : x = ln (- 1 + 2)

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} + 1 = 0

Multiply both sides by

2

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} \cdot 2 + 1 \cdot 2 = 0 \cdot 2

Simplify

e^{- x} (e^{x} + 1) (e^{x} - 1) + 2 = 0

Subtract

2

from both sides

e^{- x} (e^{x} + 1) (e^{x} - 1) + 2 - 2 = 0 - 2

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

u^{- 1} (u + 1) (u - 1) = - 2 : u = - 1 + 2, u = - 1 - 2

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

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

(u + 1) (u - 1) = - 2 u

(u + 1) (u - 1) = - 2 u

Solve

(u + 1) (u - 1) = - 2 u : u = - 1 + 2, u = - 1 - 2

(u + 1) (u - 1) = - 2 u

Expand

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

(u + 1) (u - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = u, b = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= u^{2} - 1

u^{2} - 1 = - 2 u

Move

2 u

to the left side

u^{2} - 1 = - 2 u

Add

2 u

to both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

\frac{- 2 + 2 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

= \frac{2 ( - 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

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

\frac{- 2 - 2 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 2}{2}

Factor

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

Rewrite as

= - 2 \cdot 1 - 22

Factor out common term

2

= - 2 (1 + 2)

= - \frac{2 ( 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 2)

Negate

- (1 + 2) = - 1 - 2

= - 1 - 2

The solutions to the quadratic equation are:

u = - 1 + 2, u = - 1 - 2

u = - 1 + 2, u = - 1 - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1 + 2, u = - 1 - 2

u = - 1 + 2, u = - 1 - 2

Substitute back

u = e^{x},

solve for

x

Solve

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

e^{x} = - 1 + 2

Apply exponent rules

e^{x} = - 1 + 2

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (- 1 + 2)

x = ln (- 1 + 2)

Solve

e^{x} = - 1 - 2 :

No Solution for

x \in R

e^{x} = - 1 - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (- 1 + 2)

Solve

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} - 1 = 0 : x = ln (1 + 2)

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} - 1 = 0

Multiply both sides by

2

\frac{e ^{- x} ( e ^{x} + 1 ) ( e ^{x} - 1 )}{2} \cdot 2 - 1 \cdot 2 = 0 \cdot 2

Simplify

e^{- x} (e^{x} + 1) (e^{x} - 1) - 2 = 0

Add

2

to both sides

e^{- x} (e^{x} + 1) (e^{x} - 1) - 2 + 2 = 0 + 2

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

u^{- 1} (u + 1) (u - 1) = 2 : u = 1 + 2, u = 1 - 2

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

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

(u + 1) (u - 1) = 2 u

(u + 1) (u - 1) = 2 u

Solve

(u + 1) (u - 1) = 2 u : u = 1 + 2, u = 1 - 2

(u + 1) (u - 1) = 2 u

Expand

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

(u + 1) (u - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = u, b = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= u^{2} - 1

u^{2} - 1 = 2 u

Move

2 u

to the left side

u^{2} - 1 = 2 u

Subtract

2 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

= \frac{2 ( 1 - 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

u = 1 + 2, u = 1 - 2

u = 1 + 2, u = 1 - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1 + 2, u = 1 - 2

u = 1 + 2, u = 1 - 2

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 1 + 2 : x = ln (1 + 2)

e^{x} = 1 + 2

Apply exponent rules

e^{x} = 1 + 2

f (x) = g (x)

, then

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

ln (e^{x}) = ln (1 + 2)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1 + 2)

x = ln (1 + 2)

Solve

e^{x} = 1 - 2 :

No Solution for

x \in R

e^{x} = 1 - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (1 + 2)

The solutions are

x = ln (- 1 + 2), x = ln (1 + 2)

x = ln (- 1 + 2), x = ln (1 + 2)

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Frequently Asked Questions (FAQ)

What is the general solution for 1-sinh^2(x)=0 ?
The general solution for 1-sinh^2(x)=0 is x=ln(-1+sqrt(2)),x=ln(1+sqrt(2))