You need to remember that the denominator of a fraction does not cancel, hence only numerator `sin 7x + sin 3x = 0` .

I suggest you to transform the sum `sin 7x + sin 3x` in a product such that:

`sin 7x + sin 3x = 2 sin ((7x+3x)/2)*cos ((7x-3x)/2) =gt =gt sin 7x + sin 3x = 2 sin (5x) * cos (2x)`

`` Hence, `2 sin (5x) * cos (2x) = 0` => either `sin (5x) = 0` or `cos (2x) = 0` .

You need to consider the condition `cos (9x) - cos (x) != 0` .

Transforming the difference `cos (9x) - cos (x)` in a product yields:

`cos (9x) - cos (x) = 2 sin ((9x + x)/2)* sin ((9x - x)/2)`

`cos (9x) - cos (x) = 2 sin (5x)* sin (4x)`

Hence, `2 sin (5x)* sin (4x) != 0 =gt sin (5x) != 0; sin (4x) != 0`

Since `sin (4x) != 0 lt=gt 2 sin(2x)*cos(2x) != 0 =gt cos (2x) != 0`

This condition denotes that the numerator does not cancel either.

**Hence, since the expression does not cancel => there is no solution to the equation `(sin (7x) + sin (3x))/(cos (9x) - cos (x)) = 0.` **