Question

If we have an trigonometric equation \[\tan x + 2\tan 2x + 4\tan 4x + 8\cot 8x = \sqrt 3 \] then the general solution of \[x\] is
a) \[n\pi + \dfrac{\pi }{3},\forall n \in \mathbb{Z}\]
b) \[n\pi + \dfrac{\pi }{6},\forall n \in \mathbb{Z}\]
c) \[n\pi + \dfrac{\pi }{4},\forall n \in \mathbb{Z}\]
d) \[n\pi + \pi ,\forall n \in \mathbb{Z}\]

Answer 1

Hint: We are given a trigonometric equation and we have to solve it to find the general solution of\[x\]. For this first of all we have to simplify it. We do this by adding and subtracting \[\cot x\] to the above equation. On simplifying further, we will solve the above equation by using the formula for the general solution of \[\tan x = \tan \alpha \].

Complete step-by-step solution:
We are given the equation \[\tan x + 2\tan 2x + 4\tan 4x + 8\cot 8x = \sqrt 3 \]. We have to find the general solution of \[x\]. For this first of all we add and subtract \[\cot x\] to the above equation. So,
\[\tan x - \cot x + 2\tan 2x + 4\tan 4x + 8\cot 8x + \cot x = \sqrt 3 \,\,\,\, \to (1)\]
We solve \[\tan x - \cot x\] as,
\[ \tan x - \cot x = \tan x - \dfrac{1}{{\tan x}} \\
   \Rightarrow \tan x - \cot x = \dfrac{{{{\tan }^2}x - 1}}{{\tan x}} \\ \]
We multiply and divide RHS by\[2\],
\[ \Rightarrow \tan x - \cot x = \dfrac{2}{2}\left[ {\dfrac{{{{\tan }^2}x - 1}}{{\tan x}}} \right] \\
   \Rightarrow \tan x - \cot x = - 2\left[ {\dfrac{{1 - {{\tan }^2}x}}{{2\tan x}}} \right] \\
   \Rightarrow \tan x - \cot x = \dfrac{{ - 2}}{{\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}} \]
We know that \[\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} = \tan 2x\], using this formula in the above equation,
\[ \Rightarrow \tan x - \cot x = \dfrac{{ - 2}}{{\left( {\tan 2x} \right)}} \\
   \Rightarrow \tan x - \cot x = - 2\cot 2x\,\,\, \to (2) \]
Using equation \[(2)\] in equation \[(1)\], we can say that,
\[ \Rightarrow - 2\cot 2x + 2\tan 2x + 4\tan 4x + 8\cot 8x + \cot x = \sqrt 3 \,\, \\
   \Rightarrow 2(\tan 2x - \cot 2x) + 4\tan 4x + 8\cot 8x + \cot x = \sqrt 3 \]
Using equation \[(2)\] in above step,
\[\Rightarrow - 4\cot 4x + 4\tan 4x + 8\cot 8x + \cot x = \sqrt 3 \\
   \Rightarrow 4( - \cot 4x + \tan 4x) + 8\cot 8x + \cot x = \sqrt 3 \]
Using equation \[(2)\] again, we get,
\[ \Rightarrow - 8\cot 8x + 8\cot 8x + \cot x = \sqrt 3 \\
   \Rightarrow \cot x = \sqrt 3 \\
   \Rightarrow \tan x = \dfrac{1}{{\sqrt 3 }} \]
Now we know that, general solution of the equation of form \[\tan x = \tan \alpha \] is given as,
\[n\pi \pm \alpha ,\forall n \in \mathbb{Z}\]
Here we have \[\dfrac{\pi }{6}\] in place of \[\alpha \], so we have the general solution of \[\tan x + 2\tan 2x + 4\tan 4x + 8\cot 8x = \sqrt 3 \] as \[n\pi + \dfrac{\pi }{6},\forall n \in \mathbb{Z}\].
Hence the answer is option b).

Note: We have to add and subtract or multiply and divide such that we can transform any equation in a form where we can use various formulas and identities. A general solution is one which involves the integer \[n\] and gives all solutions of a trigonometric equation. Also, the character \[\mathbb{Z}\] is used to denote the set of integers. Thus, a solution generalized by means of periodicity is known as the general solution.