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

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,

\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.