Question

Find the principal solutions of the equation $\tan x=-\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: We will first find a standard angle for which the value is given and then we will use the cofunction identities to get x. Then using the quadrants, we will find other values. Finally we will check the values that will lie between 0 to $2\pi$ and these values will be our principal solutions.
Complete step-by-step answer:
Before proceeding with the question we should understand the concept of the principal solution of trigonometric equations. The solutions lying between 0 to $2\pi$ or between ${{0}^{\circ }}$ to ${{360}^{\circ }}$ are called principal solutions.
The trigonometric equation mentioned in the question is $\tan x=-\dfrac{1}{\sqrt{3}}$.
Here tan x is negative, so x will lie in the second and fourth quadrant.
So now we know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ and also we know that $\tan (\pi -x)=-\tan x$.
Therefore, $\tan \left( \pi -\dfrac{\pi }{6} \right)=-\tan \dfrac{\pi }{6}=-\dfrac{1}{\sqrt{3}}$. So in this case x is equal to $\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$.
Also we know that $\tan (2\pi -x)=-\tan x$.
Therefore, $\tan \left( 2\pi -\dfrac{\pi }{6} \right)=-\tan \dfrac{\pi }{6}=-\dfrac{1}{\sqrt{3}}$. So in this case x is equal to $2\pi -\dfrac{\pi }{6}=\dfrac{11\pi }{6}$.
Hence both $\dfrac{5\pi }{6}$ and $\dfrac{11\pi }{6}$ lies between 0 and $2\pi$ and is known as our principal solution.
Note: We should remember that in the first quadrant the values for all trigonometric functions(sin, cos, tan, cot, sec and cosec) are positive. And in the second quadrant the values for sin and cosec are only positive. In the third quadrant the values for tan and cot are only positive. And in the fourth quadrant the values for cos and sec are only positive.