# 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}} \\

$

Answer

Verified

363.9k+ 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.

Last updated date: 01st Oct 2023

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Total views: 363.9k

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