What is the general solution for 1=sech^2(x) ?

The general solution for 1=sech^2(x) is x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1=sech^2(x)

Solution

1 = sech^{2} (x)

Solution

x = 0

Degrees

x = 0^{\circ}

Solution steps

1 = sech^{2} (x)

Switch sides

sech^{2} (x) = 1

Rewrite using trig identities

sech^{2} (x) = 1

Use the Hyperbolic identity:

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

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

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

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

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Multiply both sides by

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

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

Multiply both sides by

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

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

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

= 4 u^{2}

Simplify

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

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

Multiply:

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

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

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

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

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

Solve

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

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

Expand

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

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

(u^{2})^{2}

Apply exponent rule:

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

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

Switch sides

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

Move

4 u^{2}

to the left side

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

Subtract

4 u^{2}

from both sides

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

Simplify

u^{4} - 2 u^{2} + 1 = 0

u^{4} - 2 u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

v^{2} - 2 v + 1 = 0 : v = 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

v_{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

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

v = \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

v = 1

The solution to the quadratic equation is:

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

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 = 1, u = - 1

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1

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

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 = 0

x = 0

Popular Examples

tan(θ)=0.86667

tan (θ) = 0.86667

sin(x)*cos^2(x)=1

sin (x) \cdot cos^{2} (x) = 1

solvefor t,fw=2+cos(10pit)solve for

t, f w = 2 + cos (10 π t)

tan(θ)-sec(θ)=sqrt(3)

tan (θ) - sec (θ) = 3

2csc(x)cot(x)= 4/3

2 csc (x) cot (x) = \frac{4}{3}

Frequently Asked Questions (FAQ)

What is the general solution for 1=sech^2(x) ?
The general solution for 1=sech^2(x) is x=0