# The equation ${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ has

$

{\text{A}}{\text{. No solution}} \\

{\text{B}}{\text{. Two solutions}} \\

{\text{C}}{\text{. Three solutions}} \\

{\text{D}}{\text{. None of these}} \\

$

Last updated date: 30th Mar 2023

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Answer

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307.8k+ views

Hint:The given equation is a transcendental equation, to solve it we will make it an algebraic equation and then we will solve it using a basic algorithm approach and simplify it further.

Complete step-by-step answer:

Given equation is

$ \Rightarrow {e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ ………………………………… (1)

Multiply the equation 1 by ${e^{\sin x}}$ both sides

$ \Rightarrow {e^{2\sin x}} - 4{e^{\sin x}} - 1 = 0$………………………………… (2)

The equation (2) is a quadratic equation in terms of ${e^{\sin x}}$

Replacing ${e^{\sin x}}{\text{ by x}}$ , we get

$ \Rightarrow {x^2} - 4x - 1 = 0$ …………………………………………… (3)

Using the formula of quadratic equation to find the value of $x$

$

a{x^2} + bx + c = 0 \\

x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

$

On comparing equation (3) with the above formula, we get

\[a = 1,b = - 4,c = - 1\]

Substituting these values in the formula, we get

$

\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

\Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} \\

$

On simplifying the above equation further, we obtain

\[

\Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 4} }}{{2 \times 1}} \\

\Rightarrow x = \dfrac{{4 \pm \sqrt {20} }}{2} \\

\Rightarrow x = 2 \pm \sqrt 5 \\

\]

Now, replacing \[x\] by ${e^{\sin x}}$ , we get

$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $………………………. (4)

Converting the equation (4) from exponential to logarithmic, we get

. As we know ,$\left[ {y = {{\log }_e}(x) \Rightarrow x = {e^y}} \right]$

Since $\log (2 - \sqrt 5 )$is not defined

\[ \Rightarrow \sin x = \log (2 + \sqrt 5 )\]

As we know ,$2 + \sqrt 5 > e \Rightarrow \log (2 + \sqrt 5 ) > 1$

$ \Rightarrow \sin x > 1$ , which is not possible

Hence, for the given equation no solution exists.

Option A is correct.

Note: To solve these types of questions basic arithmetic operation and logarithmic operations are required. As we know, $y = {\log _e}(x) \Rightarrow x = {e^y}$ . There are two methods to solve quadratic equations; one is by factorising and other is using square root property. In the question, we have used the square root property as the equation is not easily factored.

Complete step-by-step answer:

Given equation is

$ \Rightarrow {e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ ………………………………… (1)

Multiply the equation 1 by ${e^{\sin x}}$ both sides

$ \Rightarrow {e^{2\sin x}} - 4{e^{\sin x}} - 1 = 0$………………………………… (2)

The equation (2) is a quadratic equation in terms of ${e^{\sin x}}$

Replacing ${e^{\sin x}}{\text{ by x}}$ , we get

$ \Rightarrow {x^2} - 4x - 1 = 0$ …………………………………………… (3)

Using the formula of quadratic equation to find the value of $x$

$

a{x^2} + bx + c = 0 \\

x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

$

On comparing equation (3) with the above formula, we get

\[a = 1,b = - 4,c = - 1\]

Substituting these values in the formula, we get

$

\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

\Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} \\

$

On simplifying the above equation further, we obtain

\[

\Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 4} }}{{2 \times 1}} \\

\Rightarrow x = \dfrac{{4 \pm \sqrt {20} }}{2} \\

\Rightarrow x = 2 \pm \sqrt 5 \\

\]

Now, replacing \[x\] by ${e^{\sin x}}$ , we get

$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $………………………. (4)

Converting the equation (4) from exponential to logarithmic, we get

. As we know ,$\left[ {y = {{\log }_e}(x) \Rightarrow x = {e^y}} \right]$

Since $\log (2 - \sqrt 5 )$is not defined

\[ \Rightarrow \sin x = \log (2 + \sqrt 5 )\]

As we know ,$2 + \sqrt 5 > e \Rightarrow \log (2 + \sqrt 5 ) > 1$

$ \Rightarrow \sin x > 1$ , which is not possible

Hence, for the given equation no solution exists.

Option A is correct.

Note: To solve these types of questions basic arithmetic operation and logarithmic operations are required. As we know, $y = {\log _e}(x) \Rightarrow x = {e^y}$ . There are two methods to solve quadratic equations; one is by factorising and other is using square root property. In the question, we have used the square root property as the equation is not easily factored.

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