
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
510k+ views
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.
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Why is steel more elastic than rubber class 11 physics CBSE

What is boron A Nonmetal B Metal C Metalloid D All class 11 chemistry CBSE

What is Environment class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

How many squares are there in a chess board A 1296 class 11 maths CBSE

