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

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

$0=-tan3x-tan4x$

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

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

$7x=npi, n in ZZ$

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

How do I solve tan9x-tan2x=tan9xtan6xtan3x+tan6xtan4xtan2xtan9x-tan2x=tan9xtan6xtan3x+tan6xtan4xtan2x?