Question

If ${{e}^{\cos x}}-{{e}^{-\cos x}}=4$ then the value of $\cos x$ will be:
A) $\log \left( 2+\sqrt{5} \right)$
B) $-\log \left( 2+\sqrt{5} \right)$
C) $\log \left( -2+\sqrt{5} \right)$
D) None of the above

Answer 1

First of all assume that ${{e}^{\cos x}}=t$ then solve the equation in terms of $t$ then compare the obtained equation from quadratic equation to find out the value of $a,b,c$ and then apply the quadratic formula to get the value of $\cos x$ and check which option is correct in the given options.

Complete step by step answer:
Trigonometry is a combination of the Greek words trigon and metron. Here, trigon refers to figures with three angles, and metron refers to measurement, hence the definition of trigonometry is triangle measurement. In today's world, trigonometry has a wide range of applications. In simple terms, we define it as the branch of mathematics concerned with angle measurement and problems relating to angles.
Trigonometry is now employed in a variety of fields, including seismology, electric circuit design, ocean tide prediction, musical tone analysis, and many others.
Trigonometric Functions are a set of real-valued functions defined as the ratio of the sides of a triangle.
The circular functions are also known as trigonometric functions.
Sine, cosine, tangent, cotangent, secant, and cosecant are the basic trigonometric functions. The sine, cosine, and tangent angles are the fundamental classifications of trigonometric functions. The cotangent, secant, and cosecant functions can all be deduced from the basic functions.
The link between triangle side lengths and angles is studied in trigonometry, a branch of mathematics. In the Hellenistic period, applications of geometry to astronomical studies gave rise to science.
Now according to the question:
Let us assume that ${{e}^{\cos x}}=t$ hence:
$\Rightarrow t-{{t}^{-1}}=4$
$\Rightarrow t-\dfrac{1}{t}=4$
Now we will take the LCM:
$\Rightarrow \dfrac{{{t}^{2}}-1}{t}=4$
Cross multiply the term we get:
$\Rightarrow {{t}^{2}}-1=4t$
Now rearrange the term:
$\Rightarrow {{t}^{2}}-4t-1=0$
Compare this equation from the quadratic equation $a{{x}^{2}}+bx+c=0$ we get $a=1,b=-4,c=-1$
Now we will apply the formula:
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Putting the values we will get:
$\Rightarrow t=\dfrac{-(-4)\pm \sqrt{{{(-4)}^{2}}-4\times 1\times (-1)}}{2\times 1}$
$\Rightarrow t=\dfrac{4\pm \sqrt{16+4}}{2}$
$\Rightarrow t=\dfrac{4\pm \sqrt{20}}{2}$
$\Rightarrow t=\dfrac{4\pm 2\sqrt{5}}{2}$
$\Rightarrow t=2\pm \sqrt{5}$
Hence we have got the values:
$\Rightarrow t=2+\sqrt{5}$ and $t=2-\sqrt{5}$
Here $t=2-\sqrt{5}$ is not possible because $2-\sqrt{5}<0$ therefore it will be eliminated as the power of the exponential term cannot be less than zero.
Hence we will use $t=2+\sqrt{5}$
Now putting the value of $t={{e}^{\cos x}}$
$\Rightarrow {{e}^{\cos x}}=2+\sqrt{5}$
$\Rightarrow \cos x={{\log }_{e}}2+\sqrt{5}$
This is not possible because the value of $2+\sqrt{5}=4.236$ but the value of $e=2.718$ and the range of $\cos \theta$ lies between $-1$ to $1$

So, the correct answer is “Option D”.

Note:
We must keep one thing in mind that $\cos \theta$ is not the same as $\cos \times \theta$ because it represents a ratio, not a product and this is true for all the trigonometric ratios. Any trigonometric function of angle ${{\theta }^{\circ }}$ is equal to the same trigonometric function of any angle $n\times {{360}^{\circ }}+\theta$, where $n$ is any integer.