What is the general solution for 2cosh^2(x)-3sinh(x)=1 ?

The general solution for 2cosh^2(x)-3sinh(x)=1 is x=ln(1.61803…),x=ln(2.41421…)

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2cosh^2(x)-3sinh(x)=1

Solution

2 cosh^{2} (x) - 3 sinh (x) = 1

Solution

x = ln (1.61803 \dots), x = ln (2.41421 \dots)

Degrees

x = 27.57140 \dots^{\circ}, x = 50.49898 \dots^{\circ}

Solution steps

2 cosh^{2} (x) - 3 sinh (x) = 1

Rewrite using trig identities

2 cosh^{2} (x) - 3 sinh (x) = 1

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

2 (\frac{e ^{x} + e ^{- x}}{2})^{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 1 : x = ln (1.61803 \dots), x = ln (2.41421 \dots)

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

2 (\frac{u + u ^{- 1}}{2})^{2} - 3 \cdot \frac{u - u ^{- 1}}{2} = 1 : u \approx - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

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

Refine

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

Multiply by LCM

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

u

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

Simplify

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

1 \cdot 2 u^{2}

Multiply the numbers:

1 \cdot 2 = 2

= 2 u^{2}

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

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

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

Solve

(u^{2} + 1)^{2} - 3 u (u^{2} - 1) = 2 u^{2} : u \approx - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

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

Expand

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 3 u, b = u^{2}, c = 1

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

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

3 u^{2} u

Apply exponent rule:

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

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

= 3 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 3 u^{3}

3 \cdot 1 \cdot u = 3 u

3 \cdot 1 \cdot u

Multiply the numbers:

3 \cdot 1 = 3

= 3 u

= - 3 u^{3} + 3 u

= - 3 u^{3} + 3 u

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

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

Move

2 u^{2}

to the left side

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

Subtract

2 u^{2}

from both sides

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

Simplify

u^{4} - 3 u^{3} + 3 u + 1 = 0

u^{4} - 3 u^{3} + 3 u + 1 = 0

Find one solution for

u^{4} - 3 u^{3} + 3 u + 1 = 0

using Newton-Raphson:

u \approx - 0.41421 \dots

u^{4} - 3 u^{3} + 3 u + 1 = 0

Newton-Raphson Approximation Definition

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

Find

f^{^{'}} (u) : 4 u^{3} - 9 u^{2} + 3

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d u} (u^{4}) - \frac{d}{d u} (3 u^{3}) + \frac{d}{d u} (3 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} (3 u^{3}) = 9 u^{2}

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

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

= 9 u^{2}

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

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

Take the constant out:

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

= 3 \frac{d u}{d u}

Apply the common derivative:

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

= 3 \cdot 1

Simplify

= 3

\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} - 9 u^{2} + 3 + 0

Simplify

= 4 u^{3} - 9 u^{2} + 3

Let

u_{0} = 0

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 0.33333 \dots : Δ u_{1} = 0.33333 \dots

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

f^{^{'}} (u_{0}) = 4 \cdot 0^{3} - 9 \cdot 0^{2} + 3 = 3

u_{1} = - 0.33333 \dots

Δ u_{1} = ∣ - 0.33333 \dots - 0 ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = - 0.4 : Δ u_{2} = 0.06666 \dots

f (u_{1}) = (- 0.33333 \dots)^{4} - 3 (- 0.33333 \dots)^{3} + 3 (- 0.33333 \dots) + 1 = 0.12345 \dots

f^{^{'}} (u_{1}) = 4 (- 0.33333 \dots)^{3} - 9 (- 0.33333 \dots)^{2} + 3 = 1.85185 \dots

u_{2} = - 0.4

Δ u_{2} = ∣ - 0.4 - (- 0.33333 \dots) ∣ = 0.06666 \dots

Δ u_{2} = 0.06666 \dots

u_{3} = - 0.41349 \dots : Δ u_{3} = 0.01349 \dots

f (u_{2}) = (- 0.4)^{4} - 3 (- 0.4)^{3} + 3 (- 0.4) + 1 = 0.0176

f^{^{'}} (u_{2}) = 4 (- 0.4)^{3} - 9 (- 0.4)^{2} + 3 = 1.304

u_{3} = - 0.41349 \dots

Δ u_{3} = ∣ - 0.41349 \dots - (- 0.4) ∣ = 0.01349 \dots

Δ u_{3} = 0.01349 \dots

u_{4} = - 0.41421 \dots : Δ u_{4} = 0.00071 \dots

f (u_{3}) = (- 0.41349 \dots)^{4} - 3 (- 0.41349 \dots)^{3} + 3 (- 0.41349 \dots) + 1 = 0.00084 \dots

f^{^{'}} (u_{3}) = 4 (- 0.41349 \dots)^{3} - 9 (- 0.41349 \dots)^{2} + 3 = 1.17838 \dots

u_{4} = - 0.41421 \dots

Δ u_{4} = ∣ - 0.41421 \dots - (- 0.41349 \dots) ∣ = 0.00071 \dots

Δ u_{4} = 0.00071 \dots

u_{5} = - 0.41421 \dots : Δ u_{5} = 2.07041 E - 6

f (u_{4}) = (- 0.41421 \dots)^{4} - 3 (- 0.41421 \dots)^{3} + 3 (- 0.41421 \dots) + 1 = 2.42567 E - 6

f^{^{'}} (u_{4}) = 4 (- 0.41421 \dots)^{3} - 9 (- 0.41421 \dots)^{2} + 3 = 1.17159 \dots

u_{5} = - 0.41421 \dots

Δ u_{5} = ∣ - 0.41421 \dots - (- 0.41421 \dots) ∣ = 2.07041 E - 6

Δ u_{5} = 2.07041 E - 6

u_{6} = - 0.41421 \dots : Δ u_{6} = 1.74062 E - 11

f (u_{5}) = (- 0.41421 \dots)^{4} - 3 (- 0.41421 \dots)^{3} + 3 (- 0.41421 \dots) + 1 = 2.03927 E - 11

f^{^{'}} (u_{5}) = 4 (- 0.41421 \dots)^{3} - 9 (- 0.41421 \dots)^{2} + 3 = 1.17157 \dots

u_{6} = - 0.41421 \dots

Δ u_{6} = ∣ - 0.41421 \dots - (- 0.41421 \dots) ∣ = 1.74062 E - 11

Δ u_{6} = 1.74062 E - 11

u \approx - 0.41421 \dots

Apply long division:

\frac{u ^{4} - 3 u ^{3} + 3 u + 1}{u + 0.41421 \dots} = u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots

u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots \approx 0

Find one solution for

u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots = 0

using Newton-Raphson:

u \approx - 0.61803 \dots

u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots = 0

Newton-Raphson Approximation Definition

f (u) = u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots

Find

f^{^{'}} (u) : 3 u^{2} - 6.82842 \dots u + 1.41421 \dots

\frac{d}{d u} (u^{3} - 3.41421 \dots u^{2} + 1.41421 \dots u + 2.41421 \dots)

Apply the Sum/Difference Rule:

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

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (3.41421 \dots u^{2}) + \frac{d}{d u} (1.41421 \dots u) + \frac{d}{d u} (2.41421 \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} (3.41421 \dots u^{2}) = 6.82842 \dots u

\frac{d}{d u} (3.41421 \dots u^{2})

Take the constant out:

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

= 3.41421 \dots \frac{d}{d u} (u^{2})

Apply the Power Rule:

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

= 3.41421 \dots \cdot 2 u^{2 - 1}

Simplify

= 6.82842 \dots u

\frac{d}{d u} (1.41421 \dots u) = 1.41421 \dots

\frac{d}{d u} (1.41421 \dots u)

Take the constant out:

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

= 1.41421 \dots \frac{d u}{d u}

Apply the common derivative:

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

= 1.41421 \dots \cdot 1

Simplify

= 1.41421 \dots

\frac{d}{d u} (2.41421 \dots) = 0

\frac{d}{d u} (2.41421 \dots)

Derivative of a constant:

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

= 0

= 3 u^{2} - 6.82842 \dots u + 1.41421 \dots + 0

Simplify

= 3 u^{2} - 6.82842 \dots u + 1.41421 \dots

Let

u_{0} = - 2

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 1.18469 \dots : Δ u_{1} = 0.81530 \dots

f (u_{0}) = (- 2)^{3} - 3.41421 \dots (- 2)^{2} + 1.41421 \dots (- 2) + 2.41421 \dots = - 22.07106 \dots

f^{^{'}} (u_{0}) = 3 (- 2)^{2} - 6.82842 \dots (- 2) + 1.41421 \dots = 27.07106 \dots

u_{1} = - 1.18469 \dots

Δ u_{1} = ∣ - 1.18469 \dots - (- 2) ∣ = 0.81530 \dots

Δ u_{1} = 0.81530 \dots

u_{2} = - 0.76792 \dots : Δ u_{2} = 0.41677 \dots

f (u_{1}) = (- 1.18469 \dots)^{3} - 3.41421 \dots (- 1.18469 \dots)^{2} + 1.41421 \dots (- 1.18469 \dots) + 2.41421 \dots = - 5.71583 \dots

f^{^{'}} (u_{1}) = 3 (- 1.18469 \dots)^{2} - 6.82842 \dots (- 1.18469 \dots) + 1.41421 \dots = 13.71437 \dots

u_{2} = - 0.76792 \dots

Δ u_{2} = ∣ - 0.76792 \dots - (- 1.18469 \dots) ∣ = 0.41677 \dots

Δ u_{2} = 0.41677 \dots

u_{3} = - 0.63287 \dots : Δ u_{3} = 0.13504 \dots

f (u_{2}) = (- 0.76792 \dots)^{3} - 3.41421 \dots (- 0.76792 \dots)^{2} + 1.41421 \dots (- 0.76792 \dots) + 2.41421 \dots = - 1.13801 \dots

f^{^{'}} (u_{2}) = 3 (- 0.76792 \dots)^{2} - 6.82842 \dots (- 0.76792 \dots) + 1.41421 \dots = 8.42703 \dots

u_{3} = - 0.63287 \dots

Δ u_{3} = ∣ - 0.63287 \dots - (- 0.76792 \dots) ∣ = 0.13504 \dots

Δ u_{3} = 0.13504 \dots

u_{4} = - 0.61820 \dots : Δ u_{4} = 0.01467 \dots

f (u_{3}) = (- 0.63287 \dots)^{3} - 3.41421 \dots (- 0.63287 \dots)^{2} + 1.41421 \dots (- 0.63287 \dots) + 2.41421 \dots = - 0.10181 \dots

f^{^{'}} (u_{3}) = 3 (- 0.63287 \dots)^{2} - 6.82842 \dots (- 0.63287 \dots) + 1.41421 \dots = 6.93738 \dots

u_{4} = - 0.61820 \dots

Δ u_{4} = ∣ - 0.61820 \dots - (- 0.63287 \dots) ∣ = 0.01467 \dots

Δ u_{4} = 0.01467 \dots

u_{5} = - 0.61803 \dots : Δ u_{5} = 0.00016 \dots

f (u_{4}) = (- 0.61820 \dots)^{3} - 3.41421 \dots (- 0.61820 \dots)^{2} + 1.41421 \dots (- 0.61820 \dots) + 2.41421 \dots = - 0.00114 \dots

f^{^{'}} (u_{4}) = 3 (- 0.61820 \dots)^{2} - 6.82842 \dots (- 0.61820 \dots) + 1.41421 \dots = 6.78208 \dots

u_{5} = - 0.61803 \dots

Δ u_{5} = ∣ - 0.61803 \dots - (- 0.61820 \dots) ∣ = 0.00016 \dots

Δ u_{5} = 0.00016 \dots

u_{6} = - 0.61803 \dots : Δ u_{6} = 2.2001 E - 8

f (u_{5}) = (- 0.61803 \dots)^{3} - 3.41421 \dots (- 0.61803 \dots)^{2} + 1.41421 \dots (- 0.61803 \dots) + 2.41421 \dots = - 1.49174 E - 7

f^{^{'}} (u_{5}) = 3 (- 0.61803 \dots)^{2} - 6.82842 \dots (- 0.61803 \dots) + 1.41421 \dots = 6.78031 \dots

u_{6} = - 0.61803 \dots

Δ u_{6} = ∣ - 0.61803 \dots - (- 0.61803 \dots) ∣ = 2.2001 E - 8

Δ u_{6} = 2.2001 E - 8

u \approx - 0.61803 \dots

Apply long division:

\frac{u ^{3} - 3.41421 \dots u ^{2} + 1.41421 \dots u + 2.41421 \dots}{u + 0.61803 \dots} = u^{2} - 4.03224 \dots u + 3.90627 \dots

u^{2} - 4.03224 \dots u + 3.90627 \dots \approx 0

Find one solution for

u^{2} - 4.03224 \dots u + 3.90627 \dots = 0

using Newton-Raphson:

u \approx 1.61803 \dots

u^{2} - 4.03224 \dots u + 3.90627 \dots = 0

Newton-Raphson Approximation Definition

f (u) = u^{2} - 4.03224 \dots u + 3.90627 \dots

Find

f^{^{'}} (u) : 2 u - 4.03224 \dots

\frac{d}{d u} (u^{2} - 4.03224 \dots u + 3.90627 \dots)

Apply the Sum/Difference Rule:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (4.03224 \dots u) + \frac{d}{d u} (3.90627 \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} (4.03224 \dots u) = 4.03224 \dots

\frac{d}{d u} (4.03224 \dots u)

Take the constant out:

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

= 4.03224 \dots \frac{d u}{d u}

Apply the common derivative:

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

= 4.03224 \dots \cdot 1

Simplify

= 4.03224 \dots

\frac{d}{d u} (3.90627 \dots) = 0

\frac{d}{d u} (3.90627 \dots)

Derivative of a constant:

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

= 0

= 2 u - 4.03224 \dots + 0

Simplify

= 2 u - 4.03224 \dots

Let

u_{0} = 1

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 1.43008 \dots : Δ u_{1} = 0.43008 \dots

f (u_{0}) = 1^{2} - 4.03224 \dots \cdot 1 + 3.90627 \dots = 0.87403 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 4.03224 \dots = - 2.03224 \dots

u_{1} = 1.43008 \dots

Δ u_{1} = ∣ 1.43008 \dots - 1 ∣ = 0.43008 \dots

Δ u_{1} = 0.43008 \dots

u_{2} = 1.58789 \dots : Δ u_{2} = 0.15781 \dots

f (u_{1}) = 1.43008 \dots^{2} - 4.03224 \dots \cdot 1.43008 \dots + 3.90627 \dots = 0.18497 \dots

f^{^{'}} (u_{1}) = 2 \cdot 1.43008 \dots - 4.03224 \dots = - 1.17208 \dots

u_{2} = 1.58789 \dots

Δ u_{2} = ∣ 1.58789 \dots - 1.43008 \dots ∣ = 0.15781 \dots

Δ u_{2} = 0.15781 \dots

u_{3} = 1.61697 \dots : Δ u_{3} = 0.02907 \dots

f (u_{2}) = 1.58789 \dots^{2} - 4.03224 \dots \cdot 1.58789 \dots + 3.90627 \dots = 0.02490 \dots

f^{^{'}} (u_{2}) = 2 \cdot 1.58789 \dots - 4.03224 \dots = - 0.85645 \dots

u_{3} = 1.61697 \dots

Δ u_{3} = ∣ 1.61697 \dots - 1.58789 \dots ∣ = 0.02907 \dots

Δ u_{3} = 0.02907 \dots

u_{4} = 1.61803 \dots : Δ u_{4} = 0.00105 \dots

f (u_{3}) = 1.61697 \dots^{2} - 4.03224 \dots \cdot 1.61697 \dots + 3.90627 \dots = 0.00084 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.61697 \dots - 4.03224 \dots = - 0.79830 \dots

u_{4} = 1.61803 \dots

Δ u_{4} = ∣ 1.61803 \dots - 1.61697 \dots ∣ = 0.00105 \dots

Δ u_{4} = 0.00105 \dots

u_{5} = 1.61803 \dots : Δ u_{5} = 1.40919 E - 6

f (u_{4}) = 1.61803 \dots^{2} - 4.03224 \dots \cdot 1.61803 \dots + 3.90627 \dots = 1.12197 E - 6

f^{^{'}} (u_{4}) = 2 \cdot 1.61803 \dots - 4.03224 \dots = - 0.79618 \dots

u_{5} = 1.61803 \dots

Δ u_{5} = ∣ 1.61803 \dots - 1.61803 \dots ∣ = 1.40919 E - 6

Δ u_{5} = 1.40919 E - 6

u_{6} = 1.61803 \dots : Δ u_{6} = 2.49326 E - 12

f (u_{5}) = 1.61803 \dots^{2} - 4.03224 \dots \cdot 1.61803 \dots + 3.90627 \dots = 1.98508 E - 12

f^{^{'}} (u_{5}) = 2 \cdot 1.61803 \dots - 4.03224 \dots = - 0.79617 \dots

u_{6} = 1.61803 \dots

Δ u_{6} = ∣ 1.61803 \dots - 1.61803 \dots ∣ = 2.49326 E - 12

Δ u_{6} = 2.49326 E - 12

u \approx 1.61803 \dots

Apply long division:

\frac{u ^{2} - 4.03224 \dots u + 3.90627 \dots}{u - 1.61803 \dots} = u - 2.41421 \dots

u - 2.41421 \dots \approx 0

u \approx 2.41421 \dots

The solutions are

u \approx - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

u \approx - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

2 (\frac{u + u ^{- 1}}{2})^{2} - 3 \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 - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

u \approx - 0.41421 \dots, u \approx - 0.61803 \dots, u \approx 1.61803 \dots, u \approx 2.41421 \dots

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = - 0.41421 \dots :

No Solution for

x \in R

e^{x} = - 0.41421 \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 :

No Solution for

x \in R

e^{x} = - 0.61803 \dots

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 1.61803 \dots : x = ln (1.61803 \dots)

e^{x} = 1.61803 \dots

Apply exponent rules

e^{x} = 1.61803 \dots

f (x) = g (x)

, then

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

ln (e^{x}) = ln (1.61803 \dots)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1.61803 \dots)

x = ln (1.61803 \dots)

Solve

e^{x} = 2.41421 \dots : x = ln (2.41421 \dots)

e^{x} = 2.41421 \dots

Apply exponent rules

e^{x} = 2.41421 \dots

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2.41421 \dots)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2.41421 \dots)

x = ln (2.41421 \dots)

x = ln (1.61803 \dots), x = ln (2.41421 \dots)

x = ln (1.61803 \dots), x = ln (2.41421 \dots)

Graph

Popular Examples

3cos^2(θ)-sin^2(θ)=2

3 cos^{2} (θ) - sin^{2} (θ) = 2

solvefor x,y= 4/pi arccos(x/4)solve for

x, y = \frac{4}{π} arccos (\frac{x}{4})

tan(θ)= 2/(1.5)

tan (θ) = \frac{2}{1.5}

tan(x)+sqrt(3)=0,0<= x<= 2pi

tan (x) + 3 = 0, 0 \leq x \leq 2 π

55=36cos((2pi)/(60)t)+39

55 = 36 cos (\frac{2 π}{60} t) + 39

Frequently Asked Questions (FAQ)

What is the general solution for 2cosh^2(x)-3sinh(x)=1 ?
The general solution for 2cosh^2(x)-3sinh(x)=1 is x=ln(1.61803…),x=ln(2.41421…)