How do you solve $7cotx-sqrt3=4cotx$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-08-06T09:55:20

The Soln. Set. $={pi/3, 4pi/3} sub [0,2pi]$ .

$7cotx-sqrt3=4cotx rArr 3cotx=sqrt3 rArr cotx=1/sqrt3$ .

$rArr tanx=sqrt3=tan(pi/3)$

$:. x=pi/3$ is a soln.

Now, the Principal Period of $tan$ function is $pi$ , i.e., to say,

$tantheta=tan(theta+pi)$ , so, $pi/3+pi=4pi/3$ is also soln.

So, the Soln. Set. $={pi/3, 4pi/3} sub [0,2pi]$ .

How do you solve 7cotx-sqrt3=4cotx7cotx-sqrt3=4cotx for 0<=x<=2pi0<=x<=2pi?