How do I solve #tan9x-tan2x=tan9xtan6xtan3x+tan6xtan4xtan2x#?

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Answer 1 · 2017-07-16T23:26:24

#x=1/7npi, n in ZZ#

#tan9x-tan2x=tan9xtan6xtan3x+tan6xtan4xtan2x# (1)

Consider #a+b=c# where #a#, #b# and #c# are all present in 1.

Then, #tan(a+b)=tanc#

#(tana+tanb)/(1-tanatanb)=tanc#

#tana+tanb=tanc(1-tanatanb)#

#tana+tanb=tanc-tanctanatanb#

#tanctanatanb=tanc-tana-tanb#

If we consider the integer triplet #a,b,c in {3, 6, 9}#, then we can derive that #tan9xtan6xtan3x=tan9x-tan6x-tan3x# (2)

If we apply the same consideration to #2,4,6# we derive #tan6xtan4xtan2x=tan6x-tan4x-tan2x# (3)

Substituting (2) and (3) into (1) yields

#cancel(tan9x)-cancel(tan2x)=cancel(tan9x)-cancel(tan6x)-tan3x+cancel(tan6x)-tan4x-cancel(tan2x)#

#0=-tan3x-tan4x#

#tan3x=-tan4x=tan(-4x)#

#3x=-4x+npi, n in ZZ#

#7x=npi, n in ZZ#

#x=1/7npi, n in ZZ#