Solve the equation #2cos^2xcotx=cotx# within the interval #0<=x<2pi#?

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Answer 1 · 2017-03-25T16:28:57

#x=npi+pi/2# or #x=npi+-pi/4# and within #0<=x<2pi#, we have #x={pi/4,pi/2,(3pi)/4,(5pi)/4,(3pi)/2,(7pi)/4}#

#2cos^2xcotx=cotx#

#hArr2cos^2xcotx-cotx=0#

or #cotx(2cos^2x-1)=0#

or #cotx(sqrt2cosx-1)(sqrt2cosx+1)=0#

Hence either #cotx=0# i.e. #x=npi+pi/2#

or #cosx=1/sqrt2=cospi/4# or #cosx=-1/sqrt2=cos((3pi)/4)#

Hence #x=npi+-pi/4#

and within #0<=x<2pi#, we have #x={pi/4,pi/2,(3pi)/4,(5pi)/4,(3pi)/2,(7pi)/4}#