Question

find the principal solution of Cot x = -√3
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Answer 1

Answer:

• Question Find the principal solution of cotx=-√3

Answer :-

$= > cot \: x = - \sqrt{3}$

$= > \frac{1}{cot \: x} = \frac{ - 1}{ \sqrt{3} }$

$= > tan \: x = \frac{ - 1}{ \sqrt{3} }$

$= > tan \: x = - tan \frac{\pi}{6}$

$= > tan \: x = (\pi - \frac{\pi}{6} )$

$= > tan \: x \: = tan \: \: \frac{5\pi}{6}$

$= > x = n\pi + \frac{5\pi}{6}$

for n belong to integers

This is the general solution of the given trionometric Q✓✓✓

To find the principal soultion we need to find x which likes at the intervel of x belong to(0-2π)

Substituting ...

when n=0. then x =

$\frac{5\pi}{6}$

Again when n=1 then x=π+ 5π/6

$x = \frac{11\pi}{6}$

$this \: lies \: in \: that \: range \because \: its \: value \:is$

33°

This two are the principal value ✓

Answer 2

${\huge {\mathfrak {Answer :-}}}$

____________

Given that :

$cot\: x = - \sqrt{3}$

Solving it :

$\implies cot\: x = - \sqrt {3}$

$\implies 1 / (cot \:x) = - 1/(\sqrt{3})$

$\implies tan \:x = - 1/(\sqrt {3})$

We know :

$tan \:30° = 1/(\sqrt{3})$

Also :

In first and third quadrant, the value of tangent is positive whereas in the second and fourth quadrant, the value is negative.

Thus :

$x$ lies either in second or in fourth quadrant.

➡️ In second quadrant,

tan 30° will become

= tan (180° - 30°) = tan 150°

➡️ In fourth quadrant,

tan 30° will become

= tan (360° - 30°) = tan 330°

.°. $x$ = 150° or 330°

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