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cosh(θ)= 7/3 \land θ<0,sinh(θ)

Solution

cosh (θ) = \frac{7}{3} and θ < 0, sinh (θ)

Solution

θ = ln (\frac{7 - 2 10}{3})

Decimal Notation

θ = - 1.49099 \dots

Solution steps

cosh (θ) = \frac{7}{3} and θ < 0

cosh (θ) = \frac{7}{3} : θ = ln (\frac{7 + 2 10}{3}), θ = ln (\frac{7 - 2 10}{3})

cosh (θ) = \frac{7}{3}

Rewrite using trig identities

cosh (θ) = \frac{7}{3}

Use the Hyperbolic identity:

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

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{7}{3}

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{7}{3}

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{7}{3} : θ = ln (\frac{7 + 2 10}{3}), θ = ln (\frac{7 - 2 10}{3})

\frac{e ^{θ} + e ^{- θ}}{2} = \frac{7}{3}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

(e^{θ} + e^{- θ}) \cdot 3 = 2 \cdot 7

Simplify

(e^{θ} + e^{- θ}) \cdot 3 = 14

Apply exponent rules

(e^{θ} + e^{- θ}) \cdot 3 = 14

Apply exponent rule:

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

e^{- θ} = (e^{θ})^{- 1}

(e^{θ} + (e^{θ})^{- 1}) \cdot 3 = 14

(e^{θ} + (e^{θ})^{- 1}) \cdot 3 = 14

Rewrite the equation with

e^{θ} = u

(u + (u)^{- 1}) \cdot 3 = 14

Solve

(u + u^{- 1}) \cdot 3 = 14 : u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

(u + u^{- 1}) \cdot 3 = 14

Refine

(u + \frac{1}{u}) \cdot 3 = 14

Simplify

(u + \frac{1}{u}) \cdot 3 : 3 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 3

Apply the commutative law:

(u + \frac{1}{u}) \cdot 3 = 3 (u + \frac{1}{u})

3 (u + \frac{1}{u})

3 (u + \frac{1}{u}) = 14

Expand

3 (u + \frac{1}{u}) : 3 u + \frac{3}{u}

3 (u + \frac{1}{u})

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = u, c = \frac{1}{u}

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

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

3 \cdot \frac{1}{u}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{u}

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

3 u + \frac{3}{u} = 14

Multiply both sides by

u

3 u + \frac{3}{u} = 14

Multiply both sides by

u

3 uu + \frac{3}{u} u = 14 u

Simplify

3 uu + \frac{3}{u} u = 14 u

Simplify

3 uu : 3 u^{2}

3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 u^{2}

Simplify

\frac{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

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

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

3 u^{2} + 3 = 14 u

3 u^{2} + 3 = 14 u

3 u^{2} + 3 = 14 u

Solve

3 u^{2} + 3 = 14 u : u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

3 u^{2} + 3 = 14 u

Move

14 u

to the left side

3 u^{2} + 3 = 14 u

Subtract

14 u

from both sides

3 u^{2} + 3 - 14 u = 14 u - 14 u

Simplify

3 u^{2} + 3 - 14 u = 0

3 u^{2} + 3 - 14 u = 0

Write in the standard form

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

3 u^{2} - 14 u + 3 = 0

Solve with the quadratic formula

3 u^{2} - 14 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 14, c = 3

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 14)^{2} - 4 \cdot 3 \cdot 3 = 410

(- 14)^{2} - 4 \cdot 3 \cdot 3

Apply exponent rule:

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

n

is even

(- 14)^{2} = 1 4^{2}

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

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 4^{2} - 36

1 4^{2} = 196

= 196 - 36

Subtract the numbers:

196 - 36 = 160

= 160

Prime factorization of

160 : 2^{5} \cdot 5

160

160

divides by

2160 = 80 \cdot 2

= 2 \cdot 80

80

divides by

280 = 40 \cdot 2

= 2 \cdot 2 \cdot 40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

= 2^{5} \cdot 5

= 2^{5} \cdot 5

Apply exponent rule:

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

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

Apply radical rule:

= 2^{4} 2 \cdot 5

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2 \cdot 5

Refine

= 410

u_{1, 2} = \frac{- ( - 14 ) \pm 4 10}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 14 ) + 4 10}{2 \cdot 3}, u_{2} = \frac{- ( - 14 ) - 4 10}{2 \cdot 3}

u = \frac{- ( - 14 ) + 4 10}{2 \cdot 3} : \frac{7 + 2 10}{3}

\frac{- ( - 14 ) + 4 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{14 + 4 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{14 + 4 10}{6}

Factor

14 + 410 : 2 (7 + 210)

14 + 410

Rewrite as

= 2 \cdot 7 + 2 \cdot 210

Factor out common term

2

= 2 (7 + 210)

= \frac{2 ( 7 + 2 10 )}{6}

Cancel the common factor:

2

= \frac{7 + 2 10}{3}

u = \frac{- ( - 14 ) - 4 10}{2 \cdot 3} : \frac{7 - 2 10}{3}

\frac{- ( - 14 ) - 4 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{14 - 4 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{14 - 4 10}{6}

Factor

14 - 410 : 2 (7 - 210)

14 - 410

Rewrite as

= 2 \cdot 7 - 2 \cdot 210

Factor out common term

2

= 2 (7 - 210)

= \frac{2 ( 7 - 2 10 )}{6}

Cancel the common factor:

2

= \frac{7 - 2 10}{3}

The solutions to the quadratic equation are:

u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u + u^{- 1}) 3

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

u = \frac{7 + 2 10}{3}, u = \frac{7 - 2 10}{3}

Substitute back

u = e^{θ},

solve for

θ

Solve

e^{θ} = \frac{7 + 2 10}{3} : θ = ln (\frac{7 + 2 10}{3})

e^{θ} = \frac{7 + 2 10}{3}

Apply exponent rules

e^{θ} = \frac{7 + 2 10}{3}

f (x) = g (x)

, then

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

ln (e^{θ}) = ln (\frac{7 + 2 10}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{θ}) = θ

θ = ln (\frac{7 + 2 10}{3})

θ = ln (\frac{7 + 2 10}{3})

Solve

e^{θ} = \frac{7 - 2 10}{3} : θ = ln (\frac{7 - 2 10}{3})

e^{θ} = \frac{7 - 2 10}{3}

Apply exponent rules

e^{θ} = \frac{7 - 2 10}{3}

f (x) = g (x)

, then

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

ln (e^{θ}) = ln (\frac{7 - 2 10}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{θ}) = θ

θ = ln (\frac{7 - 2 10}{3})

θ = ln (\frac{7 - 2 10}{3})

θ = ln (\frac{7 + 2 10}{3}), θ = ln (\frac{7 - 2 10}{3})

θ = ln (\frac{7 + 2 10}{3}), θ = ln (\frac{7 - 2 10}{3})

Combine the intervals

(θ = ln (\frac{7 - 2 10}{3}) or θ = ln (\frac{7 + 2 10}{3})) and θ < 0

Merge Overlapping Intervals

θ = ln (\frac{7 - 2 10}{3}) or θ = ln (\frac{7 + 2 10}{3}) and θ < 0

The intersection of two intervals is the set of numbers which are in both intervals

θ = ln (\frac{7 - 2 10}{3}) or θ = ln (\frac{7 + 2 10}{3})

and

θ < 0

θ = ln (\frac{7 - 2 10}{3})

θ = ln (\frac{7 - 2 10}{3})

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