What is the general solution for cosh(x)=2 ?

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

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cosh(x)=2

Solution

cosh (x) = 2

Solution

x = ln (2 + 3), x = ln (2 - 3)

Degrees

x = 75.45612 \dots^{\circ}, x = - 75.45612 \dots^{\circ}

Solution steps

cosh (x) = 2

Rewrite using trig identities

cosh (x) = 2

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

2

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

Simplify

e^{x} + e^{- x} = 4

Apply exponent rules

e^{x} + e^{- x} = 4

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

u + u^{- 1} = 4

Refine

u + \frac{1}{u} = 4

Multiply both sides by

u

u + \frac{1}{u} = 4

Multiply both sides by

u

uu + \frac{1}{u} u = 4 u

Simplify

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

u^{2} + 1 = 4 u

u^{2} + 1 = 4 u

u^{2} + 1 = 4 u

Solve

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

u^{2} + 1 = 4 u

Move

4 u

to the left side

u^{2} + 1 = 4 u

Subtract

4 u

from both sides

u^{2} + 1 - 4 u = 4 u - 4 u

Simplify

u^{2} + 1 - 4 u = 0

u^{2} + 1 - 4 u = 0

Write in the standard form

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

u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

Subtract the numbers:

16 - 4 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

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

Separate the solutions

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

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

\frac{- ( - 4 ) + 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 + 2 3}{2}

Factor

4 + 23 : 2 (2 + 3)

4 + 23

Rewrite as

= 2 \cdot 2 + 23

Factor out common term

2

= 2 (2 + 3)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 + 3

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

\frac{- ( - 4 ) - 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

The solutions to the quadratic equation are:

u = 2 + 3, u = 2 - 3

u = 2 + 3, u = 2 - 3

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u + u^{- 1}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2 + 3, u = 2 - 3

u = 2 + 3, u = 2 - 3

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 + 3 : x = ln (2 + 3)

e^{x} = 2 + 3

Apply exponent rules

e^{x} = 2 + 3

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2 + 3)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2 + 3)

x = ln (2 + 3)

Solve

e^{x} = 2 - 3 : x = ln (2 - 3)

e^{x} = 2 - 3

Apply exponent rules

e^{x} = 2 - 3

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2 - 3)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2 - 3)

x = ln (2 - 3)

x = ln (2 + 3), x = ln (2 - 3)

x = ln (2 + 3), x = ln (2 - 3)

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Frequently Asked Questions (FAQ)

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