What is the general solution for cosh(x)= 3/(sqrt(8)) ?

The general solution for cosh(x)= 3/(sqrt(8)) is x= 1/2 ln(2),x=-1/2 ln(2)

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cosh(x)= 3/(sqrt(8))

Solution

cosh (x) = \frac{3}{8}

Solution

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

Degrees

x = 19.85720 \dots^{\circ}, x = - 19.85720 \dots^{\circ}

Solution steps

cosh (x) = \frac{3}{8}

Rewrite using trig identities

cosh (x) = \frac{3}{8}

Use the Hyperbolic identity:

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

\frac{e ^{x} + e ^{- x}}{2} = \frac{3}{8}

\frac{e ^{x} + e ^{- x}}{2} = \frac{3}{8}

\frac{e ^{x} + e ^{- x}}{2} = \frac{3}{8} : x = \frac{1}{2} ln (2), x = - \frac{1}{2} ln (2)

\frac{e ^{x} + e ^{- x}}{2} = \frac{3}{8}

Apply exponent rules

\frac{e ^{x} + e ^{- x}}{2} = \frac{3}{8}

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{8} = 8^{- \frac{1}{2}}

\frac{e ^{x} + e ^{- x}}{2} = 3 \cdot 8^{- \frac{1}{2}}

\frac{e ^{x} + e ^{- x}}{2} = 3 \cdot 8^{- \frac{1}{2}}

Multiply both sides by

2

\frac{e ^{x} + e ^{- x}}{2} \cdot 2 = 3 \cdot 8^{- \frac{1}{2}} \cdot 2

Simplify

3 \cdot 8^{- \frac{1}{2}} \cdot 2 : \frac{3}{2}

3 \cdot 8^{- \frac{1}{2}} \cdot 2

8^{- \frac{1}{2}} = \frac{1}{2 2}

8^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{8}

8 = 22

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

= \frac{1}{2 2}

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2}

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Multiply by LCM

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

Find Least Common Multiplier of

u, 2 : 2 u

u, 2

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

u

2

= 2 u

Multiply by LCM=

2 u

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

Simplify

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

Simplify

u 2 u : 2 u^{2}

u 2 u

Apply exponent rule:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

Simplify

\frac{1}{u} 2 u : 2

\frac{1}{u} 2 u

Multiply fractions:

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

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

Cancel the common factor:

u

= 1 \cdot 2

Multiply:

1 \cdot 2 = 2

= 2

Simplify

\frac{3}{2} 2 u : 3 u

\frac{3}{2} 2 u

Multiply fractions:

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

= \frac{3 2}{2} u

Cancel the common factor:

2

= u \cdot 3

2 u^{2} + 2 = 3 u

2 u^{2} + 2 = 3 u

2 u^{2} + 2 = 3 u

Solve

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

2 u^{2} + 2 = 3 u

Move

3 u

to the left side

2 u^{2} + 2 = 3 u

Subtract

3 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 2

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

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

(- 3)^{2} - 422 = 1

(- 3)^{2} - 422

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

(- 3)^{2}

Apply exponent rule:

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

n

is even

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

= 3^{2}

422 = 8

422

Apply radical rule:

a a = a

22 = 2

= 4 \cdot 2

Multiply the numbers:

4 \cdot 2 = 8

= 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

= \frac{4}{2 2}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

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

Subtract the numbers:

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

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

= \frac{2}{2 2}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{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 = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = e^{x},

solve for

x

Solve

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

e^{x} = 2

Apply exponent rules

e^{x} = 2

Apply exponent rule:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

e^{x} = 2^{\frac{1}{2}}

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2^{\frac{1}{2}})

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

ln (2^{\frac{1}{2}}) = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

Solve

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

e^{x} = \frac{1}{2}

Apply exponent rules

e^{x} = \frac{1}{2}

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply exponent rule:

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

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

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

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

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

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

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

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

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

Popular Examples

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cos (x) = \frac{11}{61}

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- 1 = sec (x)

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tan (x) = \frac{59}{36}

sqrt(3)csc(θ)+2=0,0<= θ<= 2pi

3 csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

2cos^2(x)-cos(x)=3

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

Frequently Asked Questions (FAQ)

What is the general solution for cosh(x)= 3/(sqrt(8)) ?
The general solution for cosh(x)= 3/(sqrt(8)) is x= 1/2 ln(2),x=-1/2 ln(2)