How do you solve #cotx=cosx# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-08-19T03:13:11

#{pi/2, (3pi)/2}#

#cosx/sinx - cosx = 0#

#(cosxsinx - cosx)/sinx = 0#

#cosxsinx - cosx = 0#

#cosx(sinx - 1) = 0#

#cosx = 0 and sinx = 1#

#x = pi/2, (3pi)/2 and pi/2#

Note that this equation has restrictions of #x != pi, 0#, since it would render the equation undefined.

Hopefully this helps!