Tan(2x)-cot(x)=0?

Question

I solved it and got the solutions: pi/6, 5pi/6, 7pi/6 and 11pi/6. Why did I not get the pi/2 and 3pi/2 solutions?

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Answer 1 · 2018-02-18T02:01:12

My solution is found in explanation section.

#tan2x-cotx=0#

#tan2x=cotx#

#tan2x*tanx=1#

#(sin2x*sinx)/(cos2x*cosx)=1#

#[1/2*(cosx-cos3x)]/[1/2*(cos3+cosx)]=1#

#(cosx-cos3x)/(cos3x+cosx)=1#

#cosx-cos3x=cos3x+cosx#

#2cos3x=0#

#cos3x=0#

I have #2# states for #cos3x=0#,

#a)# #cos3x=cos(pi/2+2pi*k)#

#3x=pi/2+2pi*k#, so #x=pi/6+(2pi)/3*k#

Consequently, #x_1=pi/6#, #x_2=(5pi)/6# and #x_3=(3pi)/2# for #k=0, 1# and #2#

#b)# #cos3x=cos((3pi)/2+2pi*k)#

#3x=(3pi)/2+2pi*k#, so #x=pi/2+(2pi)/3*k#

Consequently, #x_4=pi/2#, #x_5=(7pi)/6# and #x_6=(11pi)/6# for #k=0, 1# and #2#