What is the general solution for e^{sin(x)}-e^{-sin(x)}=2 ?

The general solution for e^{sin(x)}-e^{-sin(x)}=2 is x=1.07876…+2pin,x=pi-1.07876…+2pin

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

Solution

e^{s i n (x)} - e^{- s i n (x)} = 2

Solution

x = 1.07876 \dots + 2 πn, x = π - 1.07876 \dots + 2 πn

Degrees

x = 61.80850 \dots^{\circ} + 36 0^{\circ} n, x = 118.19149 \dots^{\circ} + 36 0^{\circ} n

Solution steps

e^{s i n (x)} - e^{- s i n (x)} = 2

Solve by substitution

e^{s i n (x)} - e^{- s i n (x)} = 2

Let:

sin (x) = u

e^{u} - e^{- u} = 2

e^{u} - e^{- u} = 2 : u = ln (1 + 2)

e^{u} - e^{- u} = 2

Apply exponent rules

e^{u} - e^{- u} = 2

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{u} = v

v - (v)^{- 1} = 2

Solve

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

v - v^{- 1} = 2

Refine

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

Multiply both sides by

v

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

Multiply both sides by

v

vv - \frac{1}{v} v = 2 v

Simplify

vv - \frac{1}{v} v = 2 v

Simplify

vv : v^{2}

vv

Apply exponent rule:

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

vv = v^{1 + 1}

= v^{1 + 1}

Add the numbers:

1 + 1 = 2

= v^{2}

Simplify

- \frac{1}{v} v : - 1

- \frac{1}{v} v

Multiply fractions:

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

= - \frac{1 \cdot v}{v}

Cancel the common factor:

v

= - 1

v^{2} - 1 = 2 v

v^{2} - 1 = 2 v

v^{2} - 1 = 2 v

Solve

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

v^{2} - 1 = 2 v

Move

2 v

to the left side

v^{2} - 1 = 2 v

Subtract

2 v

from both sides

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

Simplify

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

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

Write in the standard form

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

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) = 22

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

Apply rule

- (- a) = a

= (- 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

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

v = 1 + 2, v = 1 - 2

v = 1 + 2, v = 1 - 2

Verify Solutions

Find undefined (singularity) points:

v = 0

Take the denominator(s) of

v - v^{- 1}

and compare to zero

v = 0

The following points are undefined

v = 0

Combine undefined points with solutions:

v = 1 + 2, v = 1 - 2

v = 1 + 2, v = 1 - 2

Substitute back

v = e^{u},

solve for

u

Solve

e^{u} = 1 + 2 : u = ln (1 + 2)

e^{u} = 1 + 2

Apply exponent rules

e^{u} = 1 + 2

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (1 + 2)

u = ln (1 + 2)

Solve

e^{u} = 1 - 2 :

No Solution for

u \in R

e^{u} = 1 - 2

a^{f (u)}

cannot be zero or negative for

u \in R

No Solution for u \in R

u = ln (1 + 2)

Substitute back

u = sin (x)

sin (x) = ln (1 + 2)

sin (x) = ln (1 + 2)

sin (x) = ln (1 + 2) : x = arcsin (ln (1 + 2)) + 2 πn, x = π - arcsin (ln (1 + 2)) + 2 πn

sin (x) = ln (1 + 2)

Apply trig inverse properties

sin (x) = ln (1 + 2)

General solutions for

sin (x) = ln (1 + 2)

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (ln (1 + 2)) + 2 πn, x = π - arcsin (ln (1 + 2)) + 2 πn

x = arcsin (ln (1 + 2)) + 2 πn, x = π - arcsin (ln (1 + 2)) + 2 πn

Combine all the solutions

x = arcsin (ln (1 + 2)) + 2 πn, x = π - arcsin (ln (1 + 2)) + 2 πn

Show solutions in decimal form

x = 1.07876 \dots + 2 πn, x = π - 1.07876 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for e^{sin(x)}-e^{-sin(x)}=2 ?
The general solution for e^{sin(x)}-e^{-sin(x)}=2 is x=1.07876…+2pin,x=pi-1.07876…+2pin