How do you find the solution to #cottheta+8=3cottheta+2# if #0<=theta<2pi#?

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Alan P. Share
Jan 10, 2017

Answer:

#theta in {0.322,3.463}# (approx.)

Explanation:

#color(blue)(cot(theta))+8=3color(blue)(cot(theta))+2#

#rarr 2color(blue)(cot(theta))=6#

#rarr color(blue)(cot(theta))=3#

If you have a calculator that evaluates #arccos# directly then
the primary value for #color(red)(theta) = color(red)(arccos(3))#
My calculator won't do that so I needed to use
#color(white)("XX")color(green)(tan(theta)=1/(cot(theta))=1/3)#
and then
#color(white)("XX")color(red)(theta)=color(red)(arctan(1/3))#
for the primary value of #color(red)(theta)#

This gave: #color(red)(theta~~0.321750554)#
as the primary value (in Quadrant 1).
Based on CAST notation for the 4 quadrants we know that a secondary value for #color(red)(theta)# will also exist in Quadrant 3 at
#color(white)("XX")color(red)(theta)~~color(red)(0.32175055+pi)=color(red)(3.463343208)#

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