Find the principal and general solutions of the equation $$cotx=-\sqrt{3}$$

Asked by Pragya Singh | 1 year ago |  92

##### Solution :-

Here it is given that,

$$cotx=-\sqrt{3}$$

Now we know that

$$cot=\dfrac{\pi}{6}=\sqrt{3}$$

And

$$cot=(\pi-\dfrac{\pi}{6})$$

$$- cot=\dfrac{\pi}{6}$$

$$-\sqrt{3}$$

$$cot=(2\pi-\dfrac{\pi}{6})=-cot\dfrac{\pi}{6}$$

=  $$-\sqrt{3}$$

Therefore we have,

$$cot\dfrac{5\pi}{6}= -\sqrt{3}$$ and $$cot\dfrac{11\pi}{6}= -\sqrt{3}$$

Therefore, the principal solutions are

$$x=\dfrac{5\pi}{6}\;and\;\dfrac{11\pi}{6}$$

And we know $$cotx= \dfrac{1}{tanx}$$

Therefore we have,

$$tanx=tan\dfrac{5\pi}{6}$$

Which implies,

$$x=n\pi \pm \dfrac{5\pi}{6}$$ , where $$n\in Z$$

Answered by Abhisek | 1 year ago

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