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# Let ${\text{S}} = \{ x \in ( - \pi ,\pi ):x \ne 0, \pm \dfrac{\pi }{2}\}$. The sum of all distinct solutions of the equation $\sqrt 3 \sec x + \cos ecx + 2(\tan x - \cot x) = 0$ in the set ${\text{S}}$ is equal to ${\text{(a) }} - \dfrac{{7\pi }}{9} \\ {\text{(b) }} - \dfrac{{2\pi }}{9} \\ {\text{(c) }}0 \\ {\text{(d) }}\dfrac{{5\pi }}{9} \\$

Last updated date: 20th Jun 2024
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Hint: For these problems, first convert the equation in the simple form and solve it. Find all the values for the variable and add them to get the final solution.

Firstly, write the expressions given in the question,
$\sqrt 3 \sec x + \cos ecx + 2(\tan x - \cot x) = 0$
Now, arrange the expression in the following way by taking the $\tan x {\text{ and }} \cot x$ to right side of equal sign,
$\Rightarrow \sqrt 3 \sec x + \cos ecx + 2(\tan x - \cot x) = 0 \\ \Rightarrow \sqrt 3 \sec x + \cos ecx = 2(\cot x - \tan x) \\$
Now, divide the equation by 2,
$\Rightarrow \dfrac{{\sqrt 3 }}{2}\sec x + \dfrac{1}{2}\cos ecx = \cot x - \tan x$
Now, convert the expression in $\sin x{\text{ and }}\cos x$ terms, take the LCM and use the formula for $\cos 2A{\text{ and }}\cos (A - B)$ in the following way,
$\Rightarrow \dfrac{{\sqrt 3 }}{2}\dfrac{1}{{\cos x}} + \dfrac{1}{2}\dfrac{1}{{\sin x}} = \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} \\ \Rightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = {\cos ^2}x - {\sin ^2}x \\ \Rightarrow \sin \dfrac{\pi }{3}\sin x + \cos \dfrac{\pi }{3}\cos x = \cos 2x \\ \Rightarrow \cos (x - \dfrac{\pi }{3}) = \cos 2x \\$
Since we know that if $\cos \theta = \cos \alpha$then$\theta = 2n\pi \pm \alpha$, hence
$\Rightarrow 2x = 2n\pi \pm (x - \dfrac{\pi }{3})$
Case 1: Consider the positive sign for the above expression then,
$\Rightarrow 2x = 2n\pi + x - \dfrac{\pi }{3} \\ \Rightarrow x = 2n\pi - \dfrac{\pi }{3} \\ {\text{If n = 0, we get }}x = - \dfrac{\pi }{3} \\ {\text{If n = 1, we get }}x = 2\pi - \dfrac{\pi }{3} = \dfrac{{5\pi }}{3} \\ {\text{If n = - 1, we get }}x = - 2\pi - \dfrac{\pi }{3} = - \dfrac{{7\pi }}{3} \\$
If you see that when either n=1 or n=-1, the values of $x$ are not in the range which is given as$x \in ( - \pi ,\pi )$. Hence both these values for $x$ will not be considered. Only one value for $x$ , i.e., $x = - \dfrac{\pi }{3}$ will be considered.
Case 1: Consider the negative sign for the above expression then,
$\Rightarrow 2x = 2n\pi - x + \dfrac{\pi }{3} \\ \Rightarrow 3x = 2n\pi + \dfrac{\pi }{3} \\ {\text{If n = 0, we get}} \\ {\text{3}}x = \dfrac{\pi }{3} \\ x = \dfrac{\pi }{9} \\ {\text{If n = 1, we get }} \\ {\text{3}}x = 2\pi + \dfrac{\pi }{3} \\ x = \dfrac{{2\pi }}{3} + \dfrac{\pi }{9} = \dfrac{{7\pi }}{9} \\ {\text{If n = - 1, we get }} \\ {\text{3}}x = - 2\pi + \dfrac{\pi }{3} \\ x = - \dfrac{{2\pi }}{3} + \dfrac{\pi }{9} = - \dfrac{{5\pi }}{9} \\ {\text{If n = - 2, we get }} \\ {\text{3}}x = - 4\pi + \dfrac{\pi }{3} \\ x = - \dfrac{{4\pi }}{3} + \dfrac{\pi }{9} = - \dfrac{{11\pi }}{9} \\$
Now, if you see that when n=-2, the value of $x$ is not in the range which is given as $x \in ( - \pi ,\pi )$. Hence both this value for $x$ will not be considered. Only 3 values for $x$ , i.e., $x = \dfrac{\pi }{9},\dfrac{{7\pi }}{9}{\text{ and }} - \dfrac{{5\pi }}{9}$will be considered.
Therefore, sum of all the distinct solutions of $x$, will be,
$x = - \dfrac{\pi }{3} + \dfrac{\pi }{9} + \dfrac{{7\pi }}{9} - \dfrac{{5\pi }}{9} \\ x = \dfrac{{ - 3\pi + \pi + 7\pi - 5\pi }}{9} \\ x = \dfrac{{ - 8\pi + 8\pi }}{9} = 0 \\$

Hence, the answer for the question is (c).

Note: For solving these questions, always remember the range for the variable because without checking from time to time, we will lose our time to reach the solution. Also, simplifying them will help to solve these questions.