You need to use the following trigonometric identity, converting the difference into a product, such that:

`sin apha - sin beta = 2cos((alpha+beta)/2)*sin((alpha-beta)/2)`

Reasoning by analogy, you need to move `sin x` to the left side, such that:

`sin(3x - 10) - sin x = 0`

Converting the difference into a product, yields:

`2cos((3x - 10 + x)/2)sin((3x - 10 - x)/2) = 0`

`2cos((4x - 10)/2)sin((2x - 10)/2) = 0`

`2cos(2x - 5)sin(x - 5) = 0`

Dividing by 2 yields:

`cos(2x - 5)sin(x - 5) = 0 => {(cos(2x - 5) = 0),(sin(x - 5) = 0):}`

`2x - 5 = cos^(-1) 0`

`2x - 5 = pi/2 => 2x = pi/2 + 5 => 2x = (pi + 10)/2 => x = pi/4 + 5/2`

`2x - 5 = (3pi)/2 => x = (3pi)/4 + 5/2`

`sin(x - 5) = 0 => x - 5 = 0 => x = 5`

`x - 5 = pi => x = pi + 5`

Since `x < 360^o = 2pi => x - 5 != 2pi`

**Hence, evaluating the solutions to the given equation yields **`x = pi/4 + 5/2 ; x = (3pi)/4 + 5/2 ; x = 5 ; x = pi + 5.`