Question

Find the general solution for $\cot x=-\sqrt{3}$

Answer 1

Hint: First we will convert cot into tan and then write that for what value of tan of the angle we get $\dfrac{-1}{\sqrt{3}}$, and then we will use the general solution of tan to find all the possible solutions, and we can see that there will be infinitely many solutions of x for which it gives $\tan x=\dfrac{-1}{\sqrt{3}}$.

Complete step-by-step answer:
Let’s convert cot into tan using the formula $\tan x=\dfrac{1}{\cot x}$
Hence, for $\cot x=-\sqrt{3}$ we get $\tan x=\dfrac{-1}{\sqrt{3}}$.
Let’s first find the value of angle for which we get $\dfrac{-1}{\sqrt{3}}$.
Now we need to find that in which quadrant tan is negative,
We know that tan is negative in ${2}^{nd}$ and ${4}^{th}$ quadrant, so $\dfrac{-\pi }{6}$ and $\pi -\dfrac{\pi }{6}$ both are the correct value,
Here, we will take $\dfrac{-\pi }{6}$.
Now we know that $\tan \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{\sqrt{3}}$
Hence, we get $\tan x=\tan \left( \dfrac{-\pi }{6} \right)$
Now we will use the formula for general solution of tan,
Now, if we have $\tan \theta =\tan \alpha $ then the general solution is:
$\theta =n\pi +\alpha $
Now using the above formula for $\tan x=\tan \left( \dfrac{-\pi }{6} \right)$ we get,
$x=n\pi -\dfrac{\pi }{6}$
Here n = integer.
Hence, from this we can see that we will get infinitely many solutions for x as we change the value of n.

Note: The formula for finding the general solution of tan is very important and must be kept in mind. In the above solution we have taken the value of $\alpha $ we have taken was $\dfrac{-\pi }{6}$, but one can also take the value of $\alpha $ as $\pi -\dfrac{\pi }{6}$ , as it lies in the ${2}^{nd}$ quadrant and gives negative value for tan. And then one can use the same formula for the general solution and replace the value of $\alpha $ with $\pi -\dfrac{\pi }{6}$ to get the answer, which is also correct.