How do I solve tan9x-tan2x=tan9xtan6xtan3x+tan6xtan4xtan2xtan9xtan2x=tan9xtan6xtan3x+tan6xtan4xtan2x?

1 Answer
Jul 16, 2017

x=1/7npi, n in ZZ

Explanation:

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