You need to convert the summation of sines into a product, such that:

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

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

Replacing `2 sin((5x)/2)*cos((-x)/2)` for `sin 2x + sin 3x` yields:

`2 sin((5x)/2)*cos((-x)/2) = (sqrt3)/2`

`sin((5x)/2)*cos((-x)/2) = ((sqrt3)/2)*(1/2)`

`cos(x/2) = 1/2 => x/2 = cos^(-1)(1/2) => x/2 = pi/3 => x = (2pi)/3`

Replacing `(2pi)/3` for x in equation `sin((5x)/2) = sqrt3/2` yields:

`sin((10pi)/6) = sin(6pi/6 + 4pi/6) => sin(pi + 2pi/3) = sin((2pi)/3)`

Using the double angle formula `sin 2alpha = 2 sin alpha*cos alpha` , yields:

`sin((2pi)/3) = 2sin(pi/3)cos(pi/3) => sin((2pi)/3) = 2(sqrt3/2)*(1/2) sin((2pi)/3) = sqrt3/2`

**Hence, testing if equation `sin((5x)/2)*cos((-x)/2) = ((sqrt3)/2)*(1/2)` holds for `x = (2pi)/3` , yields that `x = (2pi)/3` is a valid solution for equation `sin 2x + sin 3x = sqrt3/2` .**

You need to convert the summation of sines into a product, such that:

Replacing for yields:

holds for , yields that is a valid solution for equation .

NOT ALWAYS TRUE!

`sin(5pi/2)cos(-x/2)=(sqrt(3)/2)(1/2)`

`"you cannot seprate the factor, every factor is part of eauqtion!"`

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