What is the general solution for 1=sech(x) ?

The general solution for 1=sech(x) is x=0

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Popular Trigonometry >

1=sech(x)

Solution

1 = sech (x)

Solution

x = 0

Degrees

x = 0^{\circ}

Solution steps

1 = sech (x)

Switch sides

sech (x) = 1

Rewrite using trig identities

sech (x) = 1

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

e^{x} + e^{- x}

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

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

2 = u + u^{- 1}

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

2 u = u^{2} + 1

2 u = u^{2} + 1

2 u = u^{2} + 1

Solve

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

2 u = u^{2} + 1

Switch sides

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^{x},

solve for

x

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

x = 0

Verify Solutions:

x = 0

True

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Plug in

x = 0 :

True

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

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

\frac{2}{e ^{0} + e ^{- 0}}

Apply rule

a^{0} = 1, a \neq = 0

e^{0} = 1, e^{- 0} = 1

= \frac{2}{1 + 1}

Add the numbers:

1 + 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

1 = 1

True

The solution is

x = 0

x = 0

Graph

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What is the general solution for 1=sech(x) ?
The general solution for 1=sech(x) is x=0