Question

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

Answer 1

`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`