What is the general solution for cosh(x)= 3/2 ?

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

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Popular Trigonometry >

cosh(x)= 3/2

Solution

cosh (x) = \frac{3}{2}

Solution

x = ln (\frac{3 + 5}{2}), x = ln (\frac{3 - 5}{2})

Degrees

x = 55.14281 \dots^{\circ}, x = - 55.14281 \dots^{\circ}

Solution steps

cosh (x) = \frac{3}{2}

Rewrite using trig identities

cosh (x) = \frac{3}{2}

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

2

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

Simplify

e^{x} + e^{- x} = 3

Apply exponent rules

e^{x} + e^{- x} = 3

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

u + u^{- 1} = 3

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

u^{2} + 1 = 3 u

u^{2} + 1 = 3 u

Solve

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

u^{2} + 1 = 3 u

Move

3 u

to the left side

u^{2} + 1 = 3 u

Subtract

3 u

from both sides

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

Simplify

u^{2} + 1 - 3 u = 0

u^{2} + 1 - 3 u = 0

Write in the standard form

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

u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 1

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

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

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} - 4

3^{2} = 9

= 9 - 4

Subtract the numbers:

9 - 4 = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

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

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

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 5}{2}

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

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 5}{2}

The solutions to the quadratic equation are:

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

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

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

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

Substitute back

u = e^{x},

solve for

x

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{3 + 5}{2})

x = ln (\frac{3 + 5}{2})

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{3 - 5}{2})

x = ln (\frac{3 - 5}{2})

x = ln (\frac{3 + 5}{2}), x = ln (\frac{3 - 5}{2})

x = ln (\frac{3 + 5}{2}), x = ln (\frac{3 - 5}{2})

Graph

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Frequently Asked Questions (FAQ)

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