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

The general solution for 2cosh(2x)-sinh(2x)=2 is x= 1/2 ln(3),x=0

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Popular Trigonometry >

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

Solution

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

Solution

x = \frac{1}{2} ln (3), x = 0

Degrees

x = 31.47292 \dots^{\circ}, x = 0^{\circ}

Solution steps

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

Rewrite using trig identities

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

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

2 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} - \frac{e ^{2 x} - e ^{- 2 x}}{2} = 2 : x = \frac{1}{2} ln (3), x = 0

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

Multiply both sides by

2

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

Simplify

2 (e^{2 x} + e^{- 2 x}) - (e^{2 x} - e^{- 2 x}) = 4

Apply exponent rules

2 (e^{2 x} + e^{- 2 x}) - (e^{2 x} - e^{- 2 x}) = 4

Apply exponent rule:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}

2 ((e^{x})^{2} + (e^{x})^{- 2}) - ((e^{x})^{2} - (e^{x})^{- 2}) = 4

2 ((e^{x})^{2} + (e^{x})^{- 2}) - ((e^{x})^{2} - (e^{x})^{- 2}) = 4

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Simplify

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

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

Simplify

- u^{2} u^{2} : - u^{4}

- u^{2} u^{2}

Apply exponent rule:

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

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

= - u^{2 + 2}

Add the numbers:

2 + 2 = 4

= - u^{4}

Simplify

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

\frac{1}{u ^{2}} u^{2}

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1

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

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

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

Expand

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

2 (u^{2} + \frac{1}{u ^{2}}) u^{2} - u^{4} + 1

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

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = 2 u^{2}, b = u^{2}, c = \frac{1}{u ^{2}}

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

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

Simplify

2 u^{2} u^{2} + 2 \cdot \frac{1}{u ^{2}} u^{2} : 2 u^{4} + 2

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

2 u^{2} u^{2} = 2 u^{4}

2 u^{2} u^{2}

Apply exponent rule:

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

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

= 2 u^{2 + 2}

Add the numbers:

2 + 2 = 4

= 2 u^{4}

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

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

= 2 u^{4} + 2

= 2 u^{4} + 2

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

Simplify

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

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

Group like terms

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

Add similar elements:

2 u^{4} - u^{4} = u^{4}

= u^{4} + 2 + 1

Add the numbers:

2 + 1 = 3

= u^{4} + 3

= u^{4} + 3

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

Solve

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

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

Move

4 u^{2}

to the left side

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

Subtract

4 u^{2}

from both sides

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

Simplify

u^{4} + 3 - 4 u^{2} = 0

u^{4} + 3 - 4 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 4 u^{2} + 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - 4 v + 3 = 0

Solve

v^{2} - 4 v + 3 = 0 : v = 3, v = 1

v^{2} - 4 v + 3 = 0

Solve with the quadratic formula

v^{2} - 4 v + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 3

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

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

v_{1, 2} = \frac{- ( - 4 ) \pm 2}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, v_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

v = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

\frac{- ( - 4 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2}{2 \cdot 1}

Add the numbers:

4 + 2 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

v = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

\frac{- ( - 4 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2}{2 \cdot 1}

Subtract the numbers:

4 - 2 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

v = 3, v = 1

v = 3, v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3, u = - 3

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 3, u = - 3, u = 1, u = - 1

u = 3, u = - 3, u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 3, u = - 3, u = 1, u = - 1

u = 3, u = - 3, u = 1, u = - 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 3 : x = \frac{1}{2} ln (3)

e^{x} = 3

Apply exponent rules

e^{x} = 3

Apply exponent rule:

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

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

e^{x} = 3^{\frac{1}{2}}

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (3^{\frac{1}{2}})

Apply log rule:

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

ln (3^{\frac{1}{2}}) = \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3)

Solve

e^{x} = - 3 :

No Solution for

x \in R

e^{x} = - 3

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

Solve

e^{x} = - 1 :

No Solution for

x \in R

e^{x} = - 1

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = \frac{1}{2} ln (3), x = 0

x = \frac{1}{2} ln (3), x = 0

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cosh(2x)-sinh(2x)=2 ?
The general solution for 2cosh(2x)-sinh(2x)=2 is x= 1/2 ln(3),x=0