What is the general solution for csch(x)= 5/12 ?

The general solution for csch(x)= 5/12 is x=ln(5)

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Popular Trigonometry >

csch(x)= 5/12

Solution

csch (x) = \frac{5}{12}

Solution

x = ln (5)

Degrees

x = 92.21399 \dots^{\circ}

Solution steps

csch (x) = \frac{5}{12}

Rewrite using trig identities

csch (x) = \frac{5}{12}

Use the Hyperbolic identity:

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

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12} : x = ln (5)

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

2 \cdot 12 = (e^{x} - e^{- x}) \cdot 5

Simplify

24 = (e^{x} - e^{- x}) \cdot 5

Apply exponent rules

24 = (e^{x} - e^{- x}) \cdot 5

Apply exponent rule:

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

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

24 = (e^{x} - (e^{x})^{- 1}) \cdot 5

24 = (e^{x} - (e^{x})^{- 1}) \cdot 5

Rewrite the equation with

e^{x} = u

24 = (u - (u)^{- 1}) \cdot 5

Solve

24 = (u - u^{- 1}) \cdot 5 : u = 5, u = - \frac{1}{5}

24 = (u - u^{- 1}) \cdot 5

Refine

24 = (u - \frac{1}{u}) \cdot 5

Simplify

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

(u - \frac{1}{u}) \cdot 5

Apply the commutative law:

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

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

24 = 5 (u - \frac{1}{u})

Expand

5 (u - \frac{1}{u}) : 5 u - \frac{5}{u}

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

5 \cdot \frac{1}{u}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5}{u}

= 5 u - \frac{5}{u}

24 = 5 u - \frac{5}{u}

Multiply both sides by

u

24 = 5 u - \frac{5}{u}

Multiply both sides by

u

24 u = 5 uu - \frac{5}{u} u

Simplify

24 u = 5 uu - \frac{5}{u} u

Simplify

5 uu : 5 u^{2}

5 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 5 u^{2}

Simplify

- \frac{5}{u} u : - 5

- \frac{5}{u} u

Multiply fractions:

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

= - \frac{5 u}{u}

Cancel the common factor:

u

= - 5

24 u = 5 u^{2} - 5

24 u = 5 u^{2} - 5

24 u = 5 u^{2} - 5

Solve

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

24 u = 5 u^{2} - 5

Switch sides

5 u^{2} - 5 = 24 u

Move

24 u

to the left side

5 u^{2} - 5 = 24 u

Subtract

24 u

from both sides

5 u^{2} - 5 - 24 u = 24 u - 24 u

Simplify

5 u^{2} - 5 - 24 u = 0

5 u^{2} - 5 - 24 u = 0

Write in the standard form

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

5 u^{2} - 24 u - 5 = 0

Solve with the quadratic formula

5 u^{2} - 24 u - 5 = 0

Quadratic Equation Formula:

For

a = 5, b = - 24, c = - 5

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 24)^{2} - 4 \cdot 5 (- 5) = 26

(- 24)^{2} - 4 \cdot 5 (- 5)

Apply rule

- (- a) = a

= (- 24)^{2} + 4 \cdot 5 \cdot 5

Apply exponent rule:

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

n

is even

(- 24)^{2} = 2 4^{2}

= 2 4^{2} + 4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 2 4^{2} + 100

2 4^{2} = 576

= 576 + 100

Add the numbers:

576 + 100 = 676

= 676

Factor the number:

676 = 2 6^{2}

= 2 6^{2}

Apply radical rule:

n a^{n} = a

2 6^{2} = 26

= 26

u_{1, 2} = \frac{- ( - 24 ) \pm 26}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 24 ) + 26}{2 \cdot 5}, u_{2} = \frac{- ( - 24 ) - 26}{2 \cdot 5}

u = \frac{- ( - 24 ) + 26}{2 \cdot 5} : 5

\frac{- ( - 24 ) + 26}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{24 + 26}{2 \cdot 5}

Add the numbers:

24 + 26 = 50

= \frac{50}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{50}{10}

Divide the numbers:

\frac{50}{10} = 5

= 5

u = \frac{- ( - 24 ) - 26}{2 \cdot 5} : - \frac{1}{5}

\frac{- ( - 24 ) - 26}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{24 - 26}{2 \cdot 5}

Subtract the numbers:

24 - 26 = - 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 2}{10}

Apply the fraction rule:

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

= - \frac{2}{10}

Cancel the common factor:

2

= - \frac{1}{5}

The solutions to the quadratic equation are:

u = 5, u = - \frac{1}{5}

u = 5, u = - \frac{1}{5}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u - u^{- 1}) 5

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 5, u = - \frac{1}{5}

u = 5, u = - \frac{1}{5}

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 5 : x = ln (5)

e^{x} = 5

Apply exponent rules

e^{x} = 5

f (x) = g (x)

, then

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

ln (e^{x}) = ln (5)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (5)

x = ln (5)

Solve

e^{x} = - \frac{1}{5} :

No Solution for

x \in R

e^{x} = - \frac{1}{5}

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (5)

Verify Solutions:

x = ln (5)

True

Check the solutions by plugging them into

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

Remove the ones that don't agree with the equation.

Plug in

x = ln (5) :

True

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}} = \frac{5}{12}

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}} = \frac{5}{12}

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}}

e^{l n (5)} = 5

e^{l n (5)}

Apply log rule:

a^{l o g_{a} (b)} = b

= 5

e^{- l n (5)} = 5^{- 1}

e^{- l n (5)}

Apply exponent rule:

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

= (e^{l n (5)})^{- 1}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (5)} = 5

= 5^{- 1}

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

Simplify

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

Apply exponent rule:

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

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

Join

5 - \frac{1}{5} : \frac{24}{5}

5 - \frac{1}{5}

Convert element to fraction:

5 = \frac{5 \cdot 5}{5}

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

Since the denominators are equal, combine the fractions:

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

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

5 \cdot 5 - 1 = 24

5 \cdot 5 - 1

Multiply the numbers:

5 \cdot 5 = 25

= 25 - 1

Subtract the numbers:

25 - 1 = 24

= 24

= \frac{24}{5}

= \frac{2}{\frac{24}{5}}

Apply the fraction rule:

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

= \frac{2 \cdot 5}{24}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{24}

Cancel the common factor:

2

= \frac{5}{12}

= \frac{5}{12}

\frac{5}{12} = \frac{5}{12}

True

The solution is

x = ln (5)

x = ln (5)

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for csch(x)= 5/12 ?
The general solution for csch(x)= 5/12 is x=ln(5)