How do i solve this in terms of [0,2pi]? cotx-sqrt3=0

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Answer 1 · 2018-03-01T19:46:14

#color(blue)(x=pi/6 , (7pi)/6)#

Identity:

#color(red)bb(cotx=1/tanx#

#cotx-sqrt(3)=0#

Add #sqrt(3)# to both sides:

#cotx=sqrt(3)#

Using identity:

#1/tanx=sqrt(3)#

Rearranging:

#tanx=1/sqrt(3)#

#x=arctan(tanx)=arctan(1/sqrt(3))=>x=pi/6#

This is in quadrant 1. We also have an angle in quadrant III.

You can find this by:

#pi/6+pi=(7pi)/6#

So solutions are:

#color(blue)(x=pi/6 , (7pi)/6)#

Note.

For positive tangent ratios, angles lie in quadrants I and III