What is the general solution for cosh(z)=1,cosh(z)=-2 ?

The general solution for cosh(z)=1,cosh(z)=-2 is z=0

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cosh(z)=1,cosh(z)=-2

Solution

cosh (z) = 1, cosh (z) = - 2

Solution

z = 0

Degrees

z = 0^{\circ}

Solution steps

cosh (z) = 1, cosh (z) = - 2

Rewrite using trig identities

cosh (z) = 1

Use the Hyperbolic identity:

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

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

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

\frac{e ^{z} + e ^{- z}}{2} = 1 : z = 0

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

Multiply both sides by

2

\frac{e ^{z} + e ^{- z}}{2} \cdot 2 = 1 \cdot 2

Simplify

e^{z} + e^{- z} = 2

Apply exponent rules

e^{z} + e^{- z} = 2

Apply exponent rule:

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

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

e^{z} + (e^{z})^{- 1} = 2

e^{z} + (e^{z})^{- 1} = 2

Rewrite the equation with

e^{z} = u

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

Solve

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

u + u^{- 1} = 2

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

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

Cancel the common factor:

u

= 1

u^{2} + 1 = 2 u

u^{2} + 1 = 2 u

u^{2} + 1 = 2 u

Solve

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

u^{2} + 1 = 2 u

Move

2 u

to the left side

u^{2} + 1 = 2 u

Subtract

2 u

from both sides

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

Simplify

u^{2} + 1 - 2 u = 0

u^{2} + 1 - 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - 2 u + 1 = 0

Solve with the quadratic formula

u^{2} - 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u + u^{- 1}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1

u = 1

Substitute back

u = e^{z},

solve for

z

Solve

e^{z} = 1 : z = 0

e^{z} = 1

Apply exponent rules

e^{z} = 1

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{z}) = z

z = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

z = 0

z = 0

z = 0

z = 0

Popular Examples

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tan (2 x) = \frac{- 3}{3}

solvefor x,-4cos(x)=-sin^2(x)+4 solve for

x, - 4 cos (x) = - sin^{2} (x) + 4

5cos(c)+sqrt(23)=0

5 cos (c) + 23 = 0

cos(2θ)=-28/53

cos (2 θ) = - \frac{28}{53}

cos(θ)=((1))/((2))

cos (θ) = \frac{( 1 )}{( 2 )}

Frequently Asked Questions (FAQ)

What is the general solution for cosh(z)=1,cosh(z)=-2 ?
The general solution for cosh(z)=1,cosh(z)=-2 is z=0