Question

Solve the following trigonometric expression for the value of $x$ :
$\tan 2x=-\cot \left( x+\dfrac{\pi }{3} \right)$.

Answer 1

Hint: Use the formula: $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$, to change the double angle form of tangent into its single angle form. Now, change cotangent into tangent of the given angle by using, $\cot \theta =\dfrac{1}{\tan \theta }$ and break the tangent of sum of angle by using the formula: $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Simplify both the sides by cross- multiplication and then take all the terms one side. Write the expression in the form of multiplication of two terms and substitute each term equal to 0. Now, use the formula for finding the general solution of the obtained trigonometric function. If $\tan A=\tan B$, then $A=n\pi +B$.

Complete step-by-step answer:
We have been given: $\tan 2x=-\cot \left( x+\dfrac{\pi }{3} \right)$.
Using the formula: $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$, we get,
$\dfrac{2\tan x}{1-{{\tan }^{2}}x}=-\cot \left( x+\dfrac{\pi }{3} \right)$
We know that, $\cot \theta =\dfrac{1}{\tan \theta }$, therefore,
$\dfrac{2\tan x}{1-{{\tan }^{2}}x}=-\dfrac{1}{\tan \left( x+\dfrac{\pi }{3} \right)}$
Now, using the formula: $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, we get,
$\begin{aligned} & \dfrac{2\tan x}{1-{{\tan }^{2}}x}=-\dfrac{1}{\dfrac{\tan x+\tan \dfrac{\pi }{3}}{1-\tan x\tan \dfrac{\pi }{3}}} \\\ & \dfrac{2\tan x}{1-{{\tan }^{2}}x}=-\dfrac{1-\tan x\tan \dfrac{\pi }{3}}{\tan x+\tan \dfrac{\pi }{3}} \\\ \end{aligned}$
Substituting, $\tan \dfrac{\pi }{3}=\sqrt{3}$, we get,
$\dfrac{2\tan x}{1-{{\tan }^{2}}x}=-\dfrac{1-\sqrt{3}\tan x}{\tan x+\sqrt{3}}$
By cross- multiplication we get,
$\begin{aligned} & 2{{\tan }^{2}}x+2\sqrt{3}\tan x=-\left( 1-\sqrt{3}\tan x-{{\tan }^{2}}x+\sqrt{3}{{\tan }^{3}}x \right) \\\ & \Rightarrow 2{{\tan }^{2}}x+2\sqrt{3}\tan x=-1+\sqrt{3}\tan x+{{\tan }^{2}}x-\sqrt{3}{{\tan }^{3}}x \\\ & \Rightarrow 2{{\tan }^{2}}x+2\sqrt{3}\tan x+1-\sqrt{3}\tan x-{{\tan }^{2}}x+\sqrt{3}{{\tan }^{3}}x=0 \\\ & \Rightarrow {{\tan }^{2}}x+\sqrt{3}\tan x+1+\sqrt{3}{{\tan }^{3}}x=0 \\\ \end{aligned}$
The above equation can be written as:
$1+{{\tan }^{2}}x+\sqrt{3}\tan x+\sqrt{3}{{\tan }^{3}}x=0$
Taking $\sqrt{3}\tan x$ common from the last two terms, we get,
$\Rightarrow \left( 1+{{\tan }^{2}}x \right)+\sqrt{3}\tan x\left( 1+{{\tan }^{2}}x \right)=0$
Taking $\left( 1+{{\tan }^{2}}x \right)$ common, we get,
$\Rightarrow \left( 1+{{\tan }^{2}}x \right)\left( 1+\sqrt{3}\tan x \right)=0$
Substituting both the terms equal to 0, we get,

& 1+{{\tan }^{2}}x=0\text{ or }1+\sqrt{3}\tan x=0 \\\ & \Rightarrow {{\tan }^{2}}x=-1\text{ or }\sqrt{3}\tan x=-1 \\\ & \Rightarrow {{\tan }^{2}}x=-1....................(i) \\\ & \text{or }\tan x=\dfrac{-1}{\sqrt{3}}....................(ii) \\\ \end{aligned}$$ Clearly case (i) is not possible because the square of the tangent of any angle cannot be negative. Therefore, only case (ii) is possible. $\tan x=\dfrac{-1}{\sqrt{3}}$ We know that, $\tan \left( \dfrac{-\pi }{3} \right)=\dfrac{-1}{\sqrt{3}}$, therefore, $\tan x=\tan \left( \dfrac{-\pi }{3} \right)$ Now, the general solution of $\tan A=\tan B$ is given as: $A=n\pi +B$, where ‘n’ is any integer. So, the solution of above equation is: $\begin{aligned} & x=n\pi +\left( \dfrac{-\pi }{3} \right) \\\ & \Rightarrow x=n\pi -\dfrac{\pi }{3} \\\ \end{aligned}$ Note: One may note that here we have to find a general solution of the given equation because we haven’t been provided with any limits of ‘x’ to find principal solutions. We have changed all the angles into its tangent form because it is easy to solve such equations. Remember that we do not have to break the tangent form into sine and cosine form, otherwise the process of finding the solution will get complicated.