Question

How do you solve $\tan x-2-3\cot x=0$ ?

Answer 1

Hint: To solve the given equation that is a trigonometric equation as $\tan x-2-3\cot x=0$ , we will convert the whole equation in one trigonometric function. Here, we will convert the given equation in the form of $\tan x$ only. Then we do require process and calculation so that we can get the value of $x$ .

Complete step by step solution:
Since, we have the question that is a trigonometric equation as:
$\Rightarrow \tan x-2-3\cot x=0$
In the above equation, we can see that there are two trigonometric functions that are $\tan x$ and $\cot x$ . So, in the next process we will convert this equation in the form of one trigonometric function.
As we know that $\cot x$ is an inverse trigonometric function of $\tan x$ that is $\cot x=\dfrac{1}{\tan x}$ . So, we can use this rule in the above equation and can convert the above equation in the form of one trigonometric function that is $\tan x$ only here as:
$\Rightarrow \tan x-2-3\times \dfrac{1}{\tan x}=0$
Now, we will multiply by $\tan x$ both sides in the above equation. Since, left side we have $0$ . So, the result of multiplication with $0$ always gives $0$. Therefore, the above equation will be as:
$\Rightarrow \tan x\times \tan x-2\times \tan x-3\times \tan x\times \dfrac{1}{\tan x}=\tan x\times 0$
After doing the multiplication with each term in the above equation will be as:
$\Rightarrow {{\tan }^{2}}x-2\tan x-3=0$
We can write the above equation as:
$\Rightarrow {{\tan }^{2}}x-2\tan x=3$
Now, we convert it in a square by adding $1$ both sides as:
$\Rightarrow {{\tan }^{2}}x-2\tan x+1=3+1$
Since, we have the above equation in the form of ${{a}^{2}}-2ab+{{b}^{2}}$ . So we can write it as ${{\left( a-b \right)}^{2}}$ as:
$\Rightarrow {{\left( \tan x-1 \right)}^{2}}=4$
Now, taking square root both sides of the above equation we will get as:
$\Rightarrow \tan x-1=\pm 2$
Here, we will take $+2$ and $-2$ one by one as:

 Case I:
$\Rightarrow \tan x-1=+2$
Now, we will solve the equation as:
$\Rightarrow \tan x=2+1$
$\Rightarrow \tan x=3$
Here, we will take ${{\tan }^{-1}}$ both sides as:
$\Rightarrow {{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( 3 \right)$
${{\tan }^{-1}}$ will eliminate $\tan $ in the above equation as:
$\Rightarrow x={{\tan }^{-1}}\left( 3 \right)$

Case II:
$\Rightarrow \tan x-1=-2$
Now, we will solve the equation as:
$\Rightarrow \tan x=-2+1$
$\Rightarrow \tan x=-1$
Here, we will take ${{\tan }^{-1}}$ both sides as:
$\Rightarrow {{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( -1 \right)$
${{\tan }^{-1}}$ will eliminate $\tan $ in the above equation as:
$\Rightarrow x={{\tan }^{-1}}\left( -1 \right)$

Hence, the solution for the given equation $\tan x-2-3\cot x=0$ is both $x={{\tan }^{-1}}\left( 3 \right)$ and $x={{\tan }^{-1}}\left( -1 \right)$

Note: Here, we can check whether our solution is correct or not by putting any one obtained value in the following way as:
Obtain value (we will choose any one):
$\Rightarrow x={{\tan }^{-1}}\left( 3 \right)$
We will apply $\tan $ both sides as:
$\Rightarrow \tan x=\tan \left( {{\tan }^{-1}}\left( 3 \right) \right)$
Since, $\tan $ will cancel out ${{\tan }^{-1}}$ . So the above equation will be as:
$\Rightarrow \tan x=3$
As we know that $\cot x$ is equal to $\dfrac{1}{\tan x}$. So we will calculate the value of $\cot x$ as:
$\Rightarrow \cot x=\dfrac{1}{\tan x}$
$\Rightarrow \cot x=\dfrac{1}{3}$
Now, we will use the value of $\tan x$ and $\cot x$ in the given equation that is $\tan x-2-3\cot x=0$ as:
$\Rightarrow \tan x-2-3\cot x=0$
$\Rightarrow 3-2-3\times \dfrac{1}{3}=0$
Here, we will do the necessary calculation as:
$\Rightarrow 3-2-1=0$
$\Rightarrow 3-3=0$
$\Rightarrow 0=0$
Since, L.H.S. = R.H.S. Hence, the solution is correct.