What is the general solution for 2cosh^2(x)-5sinh(x)=5 ?

The general solution for 2cosh^2(x)-5sinh(x)=5 is x=ln(0.61803…),x=ln(6.16227…)

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2cosh^2(x)-5sinh(x)=5

Solution

2 cosh^{2} (x) - 5 sinh (x) = 5

Solution

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

Degrees

x = - 27.57140 \dots^{\circ}, x = 104.18930 \dots^{\circ}

Solution steps

2 cosh^{2} (x) - 5 sinh (x) = 5

Rewrite using trig identities

2 cosh^{2} (x) - 5 sinh (x) = 5

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

2 (\frac{e ^{x} + e ^{- x}}{2})^{2} - 5 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5 : x = ln (0.61803 \dots), x = ln (6.16227 \dots)

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

Apply exponent rules

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

Apply exponent rule:

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

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

2 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} - 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

2 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} - 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

Rewrite the equation with

e^{x} = u

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

Solve

2 (\frac{u + u ^{- 1}}{2})^{2} - 5 \cdot \frac{u - u ^{- 1}}{2} = 5 : u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

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

Refine

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} - \frac{5 ( u ^{2} - 1 )}{2 u} = 5

Multiply by LCM

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} - \frac{5 ( u ^{2} - 1 )}{2 u} = 5

Find Least Common Multiplier of

2 u^{2}, 2 u : 2 u^{2}

2 u^{2}, 2 u

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2 u^{2}

2 u

= 2 u^{2}

Multiply by LCM=

2 u^{2}

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

Simplify

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

Simplify

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} : (u^{2} + 1)^{2}

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2}

Multiply fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

u^{2}

= (u^{2} + 1)^{2}

Simplify

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

- \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= - \frac{5 ( u ^{2} - 1 ) u ^{2}}{u}

Cancel the common factor:

u

= - 5 u (u^{2} - 1)

Simplify

5 \cdot 2 u^{2} : 10 u^{2}

5 \cdot 2 u^{2}

Multiply the numbers:

5 \cdot 2 = 10

= 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

Solve

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2} : u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

Expand

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) : u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u

(u^{2} + 1)^{2} - 5 u (u^{2} - 1)

(u^{2} + 1)^{2} : u^{4} + 2 u^{2} + 1

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Simplify

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

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

(u^{2})^{2}

Apply exponent rule:

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

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2}

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

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

= u^{4} + 2 u^{2} + 1 - 5 u (u^{2} - 1)

Expand

- 5 u (u^{2} - 1) : - 5 u^{3} + 5 u

- 5 u (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 5 u, b = u^{2}, c = 1

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

Apply minus-plus rules

- (- a) = a

= - 5 u^{2} u + 5 \cdot 1 \cdot u

Simplify

- 5 u^{2} u + 5 \cdot 1 \cdot u : - 5 u^{3} + 5 u

- 5 u^{2} u + 5 \cdot 1 \cdot u

5 u^{2} u = 5 u^{3}

5 u^{2} u

Apply exponent rule:

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

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

= 5 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 5 u^{3}

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

Multiply the numbers:

5 \cdot 1 = 5

= 5 u

= - 5 u^{3} + 5 u

= - 5 u^{3} + 5 u

= u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u = 10 u^{2}

Move

10 u^{2}

to the left side

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u = 10 u^{2}

Subtract

10 u^{2}

from both sides

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u - 10 u^{2} = 10 u^{2} - 10 u^{2}

Simplify

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

Find one solution for

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

using Newton-Raphson:

u \approx - 0.16227 \dots

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

Newton-Raphson Approximation Definition

f (u) = u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1

Find

f^{^{'}} (u) : 4 u^{3} - 15 u^{2} - 16 u + 5

\frac{d}{d u} (u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{4}) - \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (8 u^{2}) + \frac{d}{d u} (5 u) + \frac{d}{d u} (1)

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 u^{4 - 1}

Simplify

= 4 u^{3}

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d u} (u^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 \cdot 3 u^{3 - 1}

Simplify

= 15 u^{2}

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 \cdot 2 u^{2 - 1}

Simplify

= 16 u

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 5 \cdot 1

Simplify

= 5

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 4 u^{3} - 15 u^{2} - 16 u + 5 + 0

Simplify

= 4 u^{3} - 15 u^{2} - 16 u + 5

Let

u_{0} = 0

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 0.2 : Δ u_{1} = 0.2

f (u_{0}) = 0^{4} - 5 \cdot 0^{3} - 8 \cdot 0^{2} + 5 \cdot 0 + 1 = 1

f^{^{'}} (u_{0}) = 4 \cdot 0^{3} - 15 \cdot 0^{2} - 16 \cdot 0 + 5 = 5

u_{1} = - 0.2

Δ u_{1} = ∣ - 0.2 - 0 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = - 0.16321 \dots : Δ u_{2} = 0.03678 \dots

f (u_{1}) = (- 0.2)^{4} - 5 (- 0.2)^{3} - 8 (- 0.2)^{2} + 5 (- 0.2) + 1 = - 0.2784

f^{^{'}} (u_{1}) = 4 (- 0.2)^{3} - 15 (- 0.2)^{2} - 16 (- 0.2) + 5 = 7.568

u_{2} = - 0.16321 \dots

Δ u_{2} = ∣ - 0.16321 \dots - (- 0.2) ∣ = 0.03678 \dots

Δ u_{2} = 0.03678 \dots

u_{3} = - 0.16227 \dots : Δ u_{3} = 0.00093 \dots

f (u_{2}) = (- 0.16321 \dots)^{4} - 5 (- 0.16321 \dots)^{3} - 8 (- 0.16321 \dots)^{2} + 5 (- 0.16321 \dots) + 1 = - 0.00672 \dots

f^{^{'}} (u_{2}) = 4 (- 0.16321 \dots)^{3} - 15 (- 0.16321 \dots)^{2} - 16 (- 0.16321 \dots) + 5 = 7.19444 \dots

u_{3} = - 0.16227 \dots

Δ u_{3} = ∣ - 0.16227 \dots - (- 0.16321 \dots) ∣ = 0.00093 \dots

Δ u_{3} = 0.00093 \dots

u_{4} = - 0.16227 \dots : Δ u_{4} = 6.57063 E - 7

f (u_{3}) = (- 0.16227 \dots)^{4} - 5 (- 0.16227 \dots)^{3} - 8 (- 0.16227 \dots)^{2} + 5 (- 0.16227 \dots) + 1 = - 4.72057 E - 6

f^{^{'}} (u_{3}) = 4 (- 0.16227 \dots)^{3} - 15 (- 0.16227 \dots)^{2} - 16 (- 0.16227 \dots) + 5 = 7.18434 \dots

u_{4} = - 0.16227 \dots

Δ u_{4} = ∣ - 0.16227 \dots - (- 0.16227 \dots) ∣ = 6.57063 E - 7

Δ u_{4} = 6.57063 E - 7

u \approx - 0.16227 \dots

Apply long division:

\frac{u ^{4} - 5 u ^{3} - 8 u ^{2} + 5 u + 1}{u + 0.16227 \dots} = u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots \approx 0

Find one solution for

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots = 0

using Newton-Raphson:

u \approx 0.61803 \dots

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots = 0

Newton-Raphson Approximation Definition

f (u) = u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots

Find

f^{^{'}} (u) : 3 u^{2} - 10.32455 \dots u - 7.16227 \dots

\frac{d}{d u} (u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (5.16227 \dots u^{2}) - \frac{d}{d u} (7.16227 \dots u) + \frac{d}{d u} (6.16227 \dots)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Simplify

= 3 u^{2}

\frac{d}{d u} (5.16227 \dots u^{2}) = 10.32455 \dots u

\frac{d}{d u} (5.16227 \dots u^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 5.16227 \dots \frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5.16227 \dots \cdot 2 u^{2 - 1}

Simplify

= 10.32455 \dots u

\frac{d}{d u} (7.16227 \dots u) = 7.16227 \dots

\frac{d}{d u} (7.16227 \dots u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 7.16227 \dots \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 7.16227 \dots \cdot 1

Simplify

= 7.16227 \dots

\frac{d}{d u} (6.16227 \dots) = 0

\frac{d}{d u} (6.16227 \dots)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} - 10.32455 \dots u - 7.16227 \dots + 0

Simplify

= 3 u^{2} - 10.32455 \dots u - 7.16227 \dots

Let

u_{0} = 1

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 0.64365 \dots : Δ u_{1} = 0.35634 \dots

f (u_{0}) = 1^{3} - 5.16227 \dots \cdot 1^{2} - 7.16227 \dots \cdot 1 + 6.16227 \dots = - 5.16227 \dots

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} - 10.32455 \dots \cdot 1 - 7.16227 \dots = - 14.48683 \dots

u_{1} = 0.64365 \dots

Δ u_{1} = ∣ 0.64365 \dots - 1 ∣ = 0.35634 \dots

Δ u_{1} = 0.35634 \dots

u_{2} = 0.61820 \dots : Δ u_{2} = 0.02545 \dots

f (u_{1}) = 0.64365 \dots^{3} - 5.16227 \dots \cdot 0.64365 \dots^{2} - 7.16227 \dots \cdot 0.64365 \dots + 6.16227 \dots = - 0.31981 \dots

f^{^{'}} (u_{1}) = 3 \cdot 0.64365 \dots^{2} - 10.32455 \dots \cdot 0.64365 \dots - 7.16227 \dots = - 12.56486 \dots

u_{2} = 0.61820 \dots

Δ u_{2} = ∣ 0.61820 \dots - 0.64365 \dots ∣ = 0.02545 \dots

Δ u_{2} = 0.02545 \dots

u_{3} = 0.61803 \dots : Δ u_{3} = 0.00017 \dots

f (u_{2}) = 0.61820 \dots^{3} - 5.16227 \dots \cdot 0.61820 \dots^{2} - 7.16227 \dots \cdot 0.61820 \dots + 6.16227 \dots = - 0.00210 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.61820 \dots^{2} - 10.32455 \dots \cdot 0.61820 \dots - 7.16227 \dots = - 12.39843 \dots

u_{3} = 0.61803 \dots

Δ u_{3} = ∣ 0.61803 \dots - 0.61820 \dots ∣ = 0.00017 \dots

Δ u_{3} = 0.00017 \dots

u_{4} = 0.61803 \dots : Δ u_{4} = 7.7271 E - 9

f (u_{3}) = 0.61803 \dots^{3} - 5.16227 \dots \cdot 0.61803 \dots^{2} - 7.16227 \dots \cdot 0.61803 \dots + 6.16227 \dots = - 9.57952 E - 8

f^{^{'}} (u_{3}) = 3 \cdot 0.61803 \dots^{2} - 10.32455 \dots \cdot 0.61803 \dots - 7.16227 \dots = - 12.39730 \dots

u_{4} = 0.61803 \dots

Δ u_{4} = ∣ 0.61803 \dots - 0.61803 \dots ∣ = 7.7271 E - 9

Δ u_{4} = 7.7271 E - 9

u \approx 0.61803 \dots

Apply long division:

\frac{u ^{3} - 5.16227 \dots u ^{2} - 7.16227 \dots u + 6.16227 \dots}{u - 0.61803 \dots} = u^{2} - 4.54424 \dots u - 9.97077 \dots

u^{2} - 4.54424 \dots u - 9.97077 \dots \approx 0

Find one solution for

u^{2} - 4.54424 \dots u - 9.97077 \dots = 0

using Newton-Raphson:

u \approx - 1.61803 \dots

u^{2} - 4.54424 \dots u - 9.97077 \dots = 0

Newton-Raphson Approximation Definition

f (u) = u^{2} - 4.54424 \dots u - 9.97077 \dots

Find

f^{^{'}} (u) : 2 u - 4.54424 \dots

\frac{d}{d u} (u^{2} - 4.54424 \dots u - 9.97077 \dots)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (4.54424 \dots u) - \frac{d}{d u} (9.97077 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplify

= 2 u

\frac{d}{d u} (4.54424 \dots u) = 4.54424 \dots

\frac{d}{d u} (4.54424 \dots u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4.54424 \dots \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 4.54424 \dots \cdot 1

Simplify

= 4.54424 \dots

\frac{d}{d u} (9.97077 \dots) = 0

\frac{d}{d u} (9.97077 \dots)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 2 u - 4.54424 \dots - 0

Simplify

= 2 u - 4.54424 \dots

Let

u_{0} = - 2

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 1.63510 \dots : Δ u_{1} = 0.36489 \dots

f (u_{0}) = (- 2)^{2} - 4.54424 \dots (- 2) - 9.97077 \dots = 3.11771 \dots

f^{^{'}} (u_{0}) = 2 (- 2) - 4.54424 \dots = - 8.54424 \dots

u_{1} = - 1.63510 \dots

Δ u_{1} = ∣ - 1.63510 \dots - (- 2) ∣ = 0.36489 \dots

Δ u_{1} = 0.36489 \dots

u_{2} = - 1.61807 \dots : Δ u_{2} = 0.01703 \dots

f (u_{1}) = (- 1.63510 \dots)^{2} - 4.54424 \dots (- 1.63510 \dots) - 9.97077 \dots = 0.13314 \dots

f^{^{'}} (u_{1}) = 2 (- 1.63510 \dots) - 4.54424 \dots = - 7.81446 \dots

u_{2} = - 1.61807 \dots

Δ u_{2} = ∣ - 1.61807 \dots - (- 1.63510 \dots) ∣ = 0.01703 \dots

Δ u_{2} = 0.01703 \dots

u_{3} = - 1.61803 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = (- 1.61807 \dots)^{2} - 4.54424 \dots (- 1.61807 \dots) - 9.97077 \dots = 0.00029 \dots

f^{^{'}} (u_{2}) = 2 (- 1.61807 \dots) - 4.54424 \dots = - 7.78038 \dots

u_{3} = - 1.61803 \dots

Δ u_{3} = ∣ - 1.61803 \dots - (- 1.61807 \dots) ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = - 1.61803 \dots : Δ u_{4} = 1.78938 E - 10

f (u_{3}) = (- 1.61803 \dots)^{2} - 4.54424 \dots (- 1.61803 \dots) - 9.97077 \dots = 1.3922 E - 9

f^{^{'}} (u_{3}) = 2 (- 1.61803 \dots) - 4.54424 \dots = - 7.78031 \dots

u_{4} = - 1.61803 \dots

Δ u_{4} = ∣ - 1.61803 \dots - (- 1.61803 \dots) ∣ = 1.78938 E - 10

Δ u_{4} = 1.78938 E - 10

u \approx - 1.61803 \dots

Apply long division:

\frac{u ^{2} - 4.54424 \dots u - 9.97077 \dots}{u + 1.61803 \dots} = u - 6.16227 \dots

u - 6.16227 \dots \approx 0

u \approx 6.16227 \dots

The solutions are

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

2 (\frac{u + u ^{- 1}}{2})^{2} - 5 \frac{u - u ^{- 1}}{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = - 0.16227 \dots :

No Solution for

x \in R

e^{x} = - 0.16227 \dots

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 0.61803 \dots : x = ln (0.61803 \dots)

e^{x} = 0.61803 \dots

Apply exponent rules

e^{x} = 0.61803 \dots

f (x) = g (x)

, then

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

ln (e^{x}) = ln (0.61803 \dots)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.61803 \dots)

x = ln (0.61803 \dots)

Solve

e^{x} = - 1.61803 \dots :

No Solution for

x \in R

e^{x} = - 1.61803 \dots

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 6.16227 \dots : x = ln (6.16227 \dots)

e^{x} = 6.16227 \dots

Apply exponent rules

e^{x} = 6.16227 \dots

f (x) = g (x)

, then

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

ln (e^{x}) = ln (6.16227 \dots)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (6.16227 \dots)

x = ln (6.16227 \dots)

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

Graph

Popular Examples

10+5cos(2x)=15cos(x)

10 + 5 cos (2 x) = 15 cos (x)

solvefor y,x=cos(2y)solve for

y, x = cos (2 y)

tan(x)= a/b

tan (x) = \frac{a}{b}

-1=cot(θ)

- 1 = cot (θ)

6sin^2(A)-5sin(A)+1=0

6 sin^{2} (A) - 5 sin (A) + 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cosh^2(x)-5sinh(x)=5 ?
The general solution for 2cosh^2(x)-5sinh(x)=5 is x=ln(0.61803…),x=ln(6.16227…)