You need to convert the given summation into a product, hence, you may form the following groups, such that:

`sin 3x + 2sin 2x + sin x = (sin 3x + sin x) + (sin 2x + sin 2x)`

You should notice that the term `2sin 2x` is expressed as the summation `sin 2x + sin 2x` .

The selection of terms in the two groups is decided with respect to the summation of arguments, such that:

`3x + x = 2x + 2x = 4x`

You need to convert the summations of sines into a product, using the following conversion formula, such that:

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

Reasoning by analogy, yields:

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

`sin 3x + sin x = 2sin(2x)*cos x`

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

`sin 2x + sin 2x = 2sin(2x)*cos 0`

Since `cos 0 = 1` , yields:

`sin 2x + sin 2x = 2sin(2x)`

Replacing `2sin(2x)*cos x` for `sin 3x + sin x` and `2sin(2x)` for `sin 2x + sin 2x` yields:

`sin 3x + 2sin 2x + sin x = 2sin(2x)*cos x +2sin(2x)*cos 0`

Factoring out `2sin(2x)` yields:

`sin 3x + 2sin 2x + sin x = 2sin(2x)*(cos x + cos 0)`

You need to convert the summation `cos x + cos 0` into a product, such that:

`cos x + cos 0 = 2cos((x+0)/2)*cos((x-0)/2)`

`cos x + cos 0 = 2cos^2(x/2)`

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

`sin 3x + 2sin 2x + sin x = 4sin (2x)*cos^2(x/2)`

You may use the double angle formula, such that:

`sin 3x + 2sin 2x + sin x = 4(2 sin x*cos x)*cos^2(x/2)`

`sin 3x + 2sin 2x + sin x = 8sin x*cos x*cos^2(x/2)`

**Hence, converting the given summation into a product, under the given conditions, yields **`sin 3x + 2sin 2x + sin x = 8sin x*cos x*cos^2(x/2).`

**Further Reading**

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