Question

Find the principal and the general solution of the equation $\cot x=-\sqrt{3}$

[a] General Solution is $n\pi +\dfrac{5\pi }{3}$ and the principal solution is $\dfrac{5\pi }{3},\dfrac{11\pi }{3}$ .

[b] General Solution is $n\pi -\dfrac{5\pi }{3}$ and the principal solution is $\dfrac{5\pi }{3},\dfrac{11\pi }{3}$ .

[c] General Solution is $n\pi -\dfrac{5\pi }{6}$ and the principal solution is $\dfrac{5\pi }{6},\dfrac{11\pi }{6}$ .

[d] General Solution is $n\pi +\dfrac{5\pi }{6}$ and the principal solution is $\dfrac{5\pi }{6},\dfrac{11\pi }{6}$ .

Answer 1

Hint: Period of cotx is $\pi $. Hence find the solutions in the interval $\left( 0,\pi \right)$. General solutions can be got from these solutions by adding $n\pi $. Hence if s is a solution in the interval $\left( 0,\pi \right)$ , then $n\pi +s$ is a general solution. Principal solutions are the solutions in the interval $[0,2\pi )$

Complete step-by-step answer:
We have $\cot x=-\sqrt{3}$

Since cot is positive the interval $\left( 0,\dfrac{\pi }{2} \right)$ and negative in the interval $\left( \dfrac{\pi }{2},\pi \right)$, so we have $x\in \left( \dfrac{\pi }{2},\pi \right)$

Now we know that $\cot \left( \dfrac{\pi }{6} \right)=\sqrt{3}$, hence we have $\cot \left( \pi -\dfrac{\pi }{6} \right)=\sqrt{3}$

Hence we have $x=\dfrac{5\pi }{6}$

Hence the general solution is $n\pi +\dfrac{5\pi }{6},n\in \mathbb{N}$

Put n = 0 we get $x=\dfrac{5\pi }{6}$

Put n = 1, we get $x=\dfrac{11\pi }{6}$

Hence the principal solutions are $\dfrac{5\pi }{6},\dfrac{11\pi }{6}$

Hence option [d] is correct.

Note:

[1] When solving any trigonometric equation like $\sin x=t,\cos x=t,\tan x=t,\cot x=t,\sec x=t$ or $\csc x=t$, we solve the equation first in the interval $\left[ 0,2\pi \right]$ for $\sin x=t,\cos x=t,\sec x=t$ and $\csc x=t$, in the interval $\left[ 0,\pi \right]$ for $\cot x=t$ and in the interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ for $\tan x=t$.

The general solutions for $\sin x=t,\cos x=t,\sec x=t$ and $\csc x=t$ are obtained by adding $2n\pi $ to the solutions found and for $\cot x=t$ and $\tan x=t$, the general solutions are obtained by adding $n\pi $ to the solutions found. This is because the period of functions sinx, cosx, secx and cosecx is $2\pi $ and for cotx and tanx the period is $\pi $.

[2] The mnemonic for memorizing the signs of the various trigonometric functions in various quadrants is “Add Sugar To Coffee”.

In the first quadrant, All are positive
In the second quadrant, sinx and cosecx (inverse of sinx) are positive. Rest all are negative.
In the third quadrant, tanx and cotx (inverse of tanx) are positive. Rest all are negative.
In the fourth quadrant, cosx and secx (inverse of cosx) are positive. Rest all are negative.