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

The general solution for 3cosh(2x)=5 is x= 1/2 ln(3),x=-1/2 ln(3)

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Popular Trigonometry >

3cosh(2x)=5

Solution

3 cosh (2 x) = 5

Solution

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

Degrees

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

Solution steps

3 cosh (2 x) = 5

Rewrite using trig identities

3 cosh (2 x) = 5

Use the Hyperbolic identity:

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

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

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

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

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

3 (u^{4} + 1) = 10 u^{2}

3 (u^{4} + 1) = 10 u^{2}

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

3 u^{4} + 3 = 10 u^{2}

Move

10 u^{2}

to the left side

3 u^{4} + 3 = 10 u^{2}

Subtract

10 u^{2}

from both sides

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

Simplify

3 u^{4} + 3 - 10 u^{2} = 0

3 u^{4} + 3 - 10 u^{2} = 0

Write in the standard form

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

3 u^{4} - 10 u^{2} + 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

3 v^{2} - 10 v + 3 = 0

Solve

3 v^{2} - 10 v + 3 = 0 : v = 3, v = \frac{1}{3}

3 v^{2} - 10 v + 3 = 0

Solve with the quadratic formula

3 v^{2} - 10 v + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 10, c = 3

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

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

(- 10)^{2} - 4 \cdot 3 \cdot 3 = 8

(- 10)^{2} - 4 \cdot 3 \cdot 3

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

Subtract the numbers:

100 - 36 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 3}, v_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 3}

v = \frac{- ( - 10 ) + 8}{2 \cdot 3} : 3

\frac{- ( - 10 ) + 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{10 + 8}{2 \cdot 3}

Add the numbers:

10 + 8 = 18

= \frac{18}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= 3

v = \frac{- ( - 10 ) - 8}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 10 ) - 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{10 - 8}{2 \cdot 3}

Subtract the numbers:

10 - 8 = 2

= \frac{2}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

The solutions to the quadratic equation are:

v = 3, v = \frac{1}{3}

v = 3, v = \frac{1}{3}

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} = \frac{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

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

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

\frac{1}{3} = \frac{1}{3}

\frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{3}

- \frac{1}{3} = - \frac{1}{3}

- \frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

The solutions are

u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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 = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

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

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

Apply exponent rules

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply exponent rule:

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

No Solution for

x \in R

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

Apply exponent rules

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

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

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cosh(2x)=5 ?
The general solution for 3cosh(2x)=5 is x= 1/2 ln(3),x=-1/2 ln(3)