You need to convert the products from both sides into summations, using the following trigonometric identity, such that:

`sin a*sin b = (cos(a - b) - cos(a + b))/2`

Reasoning by analogy yields:

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

`sin x*sin 5x = (cos(-4x) - cos(6x))/2`

Since `cos(-x) = cos x` yields:

`sin x*sin 5x = (cos(4x) - cos(6x))/2`

`sin 2x*sin 6x = (cos(2x - 6x) - cos(2x + 6x))/2`

`sin 2x*sin 6x = (cos(4x) - cos(8x))/2`

Replacing `(cos(4x) - cos(6x))/2` for `sin x*sin 5x` and `(cos(4x) - cos(8x))/2` for `sin 2x*sin 6x` yields:

`(cos(4x) - cos(6x))/2 = (cos(4x) - cos(8x))/2`

Reducing duplicate factors yields:

`cos(4x) - cos(6x) = cos(4x) - cos(8x)`

`- cos(6x) = - cos(8x) => cos(6x) = cos(8x)`

`cos(6x) - cos(8x) = 0`

Converting the difference into a product yields:

`cos(6x) - cos(8x) = 2sin(3x + 4x)sin(4x - 3x)`

`cos(6x) - cos(8x) = 2sin 7x*sin x`

Using zero product rule yields:

`2sin 7x*sin x = 0 => {(sin 7x = 0),(sin x = 0):}`

`{(7x = (-1)^n*sin^(-1)0 + n*pi),(x = (-1)^n*sin^(-1)0 + n*pi):}`

`{(x = (n*pi)/7),(x = n*pi):}`

**Hence, evaluating the general solutions to the given equation, yields **`x = n*pi, x = (n*pi)/7.`