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Question: The equation \({e^{\sin x}} - {e^{ - \sin x}} - 4 = 0\) has \( {\text{A}}{\text{. No solution}} ...

The equation esin⁡x−e−sin⁡x−4=0{e^{\sin x}} - {e^{ - \sin x}} - 4 = 0 has
A. No solution B. Two solutions C. Three solutions D. None of these  {\text{A}}{\text{. No solution}} \\\ {\text{B}}{\text{. Two solutions}} \\\ {\text{C}}{\text{. Three solutions}} \\\ {\text{D}}{\text{. None of these}} \\\

Explanation

Solution

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
⇒esin⁡x−e−sin⁡x−4=0\Rightarrow {e^{\sin x}} - {e^{ - \sin x}} - 4 = 0 ………………………………… (1)
Multiply the equation 1 by esin⁡x{e^{\sin x}} both sides
⇒e2sin⁡x−4esin⁡x−1=0\Rightarrow {e^{2\sin x}} - 4{e^{\sin x}} - 1 = 0………………………………… (2)
The equation (2) is a quadratic equation in terms of esin⁡x{e^{\sin x}}
Replacing esin⁡x by x{e^{\sin x}}{\text{ by x}} , we get
⇒x2−4x−1=0\Rightarrow {x^2} - 4x - 1 = 0 …………………………………………… (3)
Using the formula of quadratic equation to find the value of xx
ax2+bx+c=0 x=−b±b2−4ac2a  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=−1a = 1,b = - 4,c = - 1

Substituting these values in the formula, we get

⇒x=−b±b2−4ac2a ⇒x=−(−4)±(−4)2−4×1×(−1)2×1  \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

⇒x=4±16+42×1 ⇒x=4±202 ⇒x=2±5 \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 xx by esin⁡x{e^{\sin x}} , we get
⇒esin⁡x=2±5\Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5………………………. (4)
Converting the equation (4) from exponential to logarithmic, we get
. As we know ,[y=log⁡e(x)⇒x=ey]\left[ {y = {{\log }_e}(x) \Rightarrow x = {e^y}} \right]
Since log⁡(2−5)\log (2 - \sqrt 5 )is not defined
⇒sin⁡x=log⁡(2+5)\Rightarrow \sin x = \log (2 + \sqrt 5 )
As we know ,2+5>e⇒log⁡(2+5)>12 + \sqrt 5 > e \Rightarrow \log (2 + \sqrt 5 ) > 1
⇒sin⁡x>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)⇒x=eyy = {\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.