What is the general solution for 5sinh(2x)+3cosh(2x)=4 ?

The general solution for 5sinh(2x)+3cosh(2x)=4 is x= 1/2 ln((1+sqrt(2))/2)

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Popular Trigonometry >

5sinh(2x)+3cosh(2x)=4

Solution

5 sinh (2 x) + 3 cosh (2 x) = 4

Solution

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

Degrees

x = 5.39228 \dots^{\circ}

Solution steps

5 sinh (2 x) + 3 cosh (2 x) = 4

Rewrite using trig identities

5 sinh (2 x) + 3 cosh (2 x) = 4

Use the Hyperbolic identity:

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

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 cosh (2 x) = 4

Use the Hyperbolic identity:

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

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4 : x = \frac{1}{2} ln (\frac{1 + 2}{2})

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

Apply exponent rules

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

Apply exponent rule:

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

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

5 \cdot \frac{( e ^{x} ) ^{2} - ( e ^{x} ) ^{- 2}}{2} + 3 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} = 4

5 \cdot \frac{( e ^{x} ) ^{2} - ( e ^{x} ) ^{- 2}}{2} + 3 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} = 4

Rewrite the equation with

e^{x} = u

5 \cdot \frac{( u ) ^{2} - ( u ) ^{- 2}}{2} + 3 \cdot \frac{( u ) ^{2} + ( u ) ^{- 2}}{2} = 4

Solve

5 \cdot \frac{u ^{2} - u ^{- 2}}{2} + 3 \cdot \frac{u ^{2} + u ^{- 2}}{2} = 4 : u = \frac{1 + 2}{2}, u = - \frac{1 + 2}{2}

5 \cdot \frac{u ^{2} - u ^{- 2}}{2} + 3 \cdot \frac{u ^{2} + u ^{- 2}}{2} = 4

Refine

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

Multiply both sides by

2 u^{2}

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

Multiply both sides by

2 u^{2}

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

Simplify

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

Simplify

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2} : 5 (u^{4} - 1)

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2}

Multiply fractions:

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

= \frac{5 ( u ^{4} - 1 ) \cdot 2 u ^{2}}{2 u ^{2}}

Cancel the common factor:

2

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

Cancel the common factor:

u^{2}

= 5 (u^{4} - 1)

Simplify

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

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

Multiply fractions:

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

= \frac{3 ( u ^{4} + 1 ) \cdot 2 u ^{2}}{2 u ^{2}}

Cancel the common factor:

2

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

Cancel the common factor:

u^{2}

= 3 (u^{4} + 1)

Simplify

4 \cdot 2 u^{2} : 8 u^{2}

4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 2 = 8

= 8 u^{2}

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

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

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

Solve

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

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

Expand

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

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

Expand

5 (u^{4} - 1) : 5 u^{4} - 5

5 (u^{4} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = u^{4}, c = 1

= 5 u^{4} - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 u^{4} - 5

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

Expand

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

3 (u^{4} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = u^{4}, c = 1

= 3 u^{4} + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 u^{4} + 3

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

Simplify

5 u^{4} - 5 + 3 u^{4} + 3 : 8 u^{4} - 2

5 u^{4} - 5 + 3 u^{4} + 3

Group like terms

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

Add similar elements:

5 u^{4} + 3 u^{4} = 8 u^{4}

= 8 u^{4} - 5 + 3

Add/Subtract the numbers:

- 5 + 3 = - 2

= 8 u^{4} - 2

= 8 u^{4} - 2

8 u^{4} - 2 = 8 u^{2}

Move

8 u^{2}

to the left side

8 u^{4} - 2 = 8 u^{2}

Subtract

8 u^{2}

from both sides

8 u^{4} - 2 - 8 u^{2} = 8 u^{2} - 8 u^{2}

Simplify

8 u^{4} - 2 - 8 u^{2} = 0

8 u^{4} - 2 - 8 u^{2} = 0

Write in the standard form

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

8 u^{4} - 8 u^{2} - 2 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

8 v^{2} - 8 v - 2 = 0

Solve

8 v^{2} - 8 v - 2 = 0 : v = \frac{1 + 2}{2}, v = \frac{1 - 2}{2}

8 v^{2} - 8 v - 2 = 0

Solve with the quadratic formula

8 v^{2} - 8 v - 2 = 0

Quadratic Equation Formula:

For

a = 8, b = - 8, c = - 2

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

(- 8)^{2} - 4 \cdot 8 (- 2) = 82

(- 8)^{2} - 4 \cdot 8 (- 2)

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 8 \cdot 2

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 8 \cdot 2

Multiply the numbers:

4 \cdot 8 \cdot 2 = 64

= 8^{2} + 64

8^{2} = 64

= 64 + 64

Add the numbers:

64 + 64 = 128

= 128

Prime factorization of

128 : 2^{7}

128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

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

2

is a prime number, therefore no further factorization is possible

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

= 2^{7}

= 2^{7}

Apply exponent rule:

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

= 2^{6} \cdot 2

Apply radical rule:

= 2 2^{6}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 2

Refine

= 82

v_{1, 2} = \frac{- ( - 8 ) \pm 8 2}{2 \cdot 8}

Separate the solutions

v_{1} = \frac{- ( - 8 ) + 8 2}{2 \cdot 8}, v_{2} = \frac{- ( - 8 ) - 8 2}{2 \cdot 8}

v = \frac{- ( - 8 ) + 8 2}{2 \cdot 8} : \frac{1 + 2}{2}

\frac{- ( - 8 ) + 8 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{8 + 8 2}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8 + 8 2}{16}

Factor

8 + 82 : 8 (1 + 2)

8 + 82

Rewrite as

= 8 \cdot 1 + 82

Factor out common term

8

= 8 (1 + 2)

= \frac{8 ( 1 + 2 )}{16}

Cancel the common factor:

8

= \frac{1 + 2}{2}

v = \frac{- ( - 8 ) - 8 2}{2 \cdot 8} : \frac{1 - 2}{2}

\frac{- ( - 8 ) - 8 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{8 - 8 2}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8 - 8 2}{16}

Factor

8 - 82 : 8 (1 - 2)

8 - 82

Rewrite as

= 8 \cdot 1 - 82

Factor out common term

8

= 8 (1 - 2)

= \frac{8 ( 1 - 2 )}{16}

Cancel the common factor:

8

= \frac{1 - 2}{2}

The solutions to the quadratic equation are:

v = \frac{1 + 2}{2}, v = \frac{1 - 2}{2}

v = \frac{1 + 2}{2}, v = \frac{1 - 2}{2}

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

Solve

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

No Solution for

u \in R

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

x^{2}

cannot be negative for

x \in R

No Solution for u \in R

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

5 \frac{u ^{2} - u ^{- 2}}{2} + 3 \frac{u ^{2} + u ^{- 2}}{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 = \frac{1 + 2}{2}, u = - \frac{1 + 2}{2}

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

Substitute back

u = e^{x},

solve for

x

Solve

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

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

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

Apply log rule:

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

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

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

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

Solve

e^{x} = - \frac{1 + 2}{2} :

No Solution for

x \in R

e^{x} = - \frac{1 + 2}{2}

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

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

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

Graph

Popular Examples

3sin(2t)+4=1,-pi/2 <= t<= pi/2

Frequently Asked Questions (FAQ)

What is the general solution for 5sinh(2x)+3cosh(2x)=4 ?
The general solution for 5sinh(2x)+3cosh(2x)=4 is x= 1/2 ln((1+sqrt(2))/2)