# Show that equation sin7x+sin3x/cos9x-cosx=0 hasn't solution?

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### 1 Answer

You need to transform the sum `sin 7x + sin 3x` and difference `cos 9x - cos x` into products such that:

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

`sin 7x + sin 3x =2 sin 5x*cos 2x`

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

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

The fraction `(sin7x+sin3x)/(cos9x-cosx) ` cancels only if `sin7x+sin3x = 0` , hence `2 sin 5x*cos 2x = 0 =gt sin 5x = 0` or `cos 2x = 0`

Notice that `sin 5x` is factor of product `-2 sin 5x* sin 4x` found to denominator. Since denominator needs to be different of 0, then `sin 5x != 0` , which contradicts `sin 5x= 0` .

Considering `sin 5x!=0` , then only `cos 2x = 0` may cancel the numerator.

You notice that `cos 2x` is also a factor of product to denominator since `sin 4x = 2 sin 2x* cos 2x` , hence `cos 2x!= 0` .

Since `sin 5x ` and `cos 2x` are both factors of products found to numerator and denominator, they need to be different of 0.

**Since `sin 5x!=0` ; `cos 2x!=0` => `2 sin 5x*cos 2x != 0` , hence the fraction `(sin7x+sin3x)/(cos9x-cosx) != 0` and there are no solutions to equation `(sin7x+sin3x)/(cos9x-cosx) = ` 0.**