# How do I solve Sin2x + Sin3x = squareroot(3)/2 for domain 0 to piThis is a Year 12 question

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### 1 Answer

You need to convert the sum of sines in 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)`

Since the cosine function is even, then `cos (-x/2) = cos x/2`

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

You need to divide by 2 both sides such that:

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

You need to consider the following possibilities such that:

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

or

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

If `sin (5x/2) = sqrt3/2` and `cos (x/2) = 1/2` for `x in [0,pi]` , then `5x/2 = pi/3 ` or `5x/2 = pi - pi/3 = 2pi/3` .

Since, both values of sine and cosine function are positive for `x in [0,pi/2]` , then you need to consider only `5x/2=pi/3` and `x/2 = pi/6` .

`5x/2 = pi/3 =gt x = 2pi/15`

`cos x/2 = 1/2 =gt x/2 = pi/6 =gt x = pi/3`

If `sin (5x/2) = 1/2` and `cos (x/2) = sqrt3/2` for `x in [0,pi], ` then `5x/2 = pi/6` and`x/2 = pi/3` .

`x = pi/15 and x = 2pi/3`

**Hence, evaluating the values of `x in [0,pi]` yields `pi/15, 2pi/15, pi/3, 2pi/3.` **