# Sin2xcos2x

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### 1 Answer

The problem does not specify its requests, hence, supposing that you need to solve for x the equation `sin 2x*cos 2x = 0` , you may perform the following steps:

- multiply both sides by 2, such that:

`2sin 2x*cos 2x = 2*0`

- use the double angle identity `2sin theta*cos theta = sin 2theta` , such that:

`sin 2*(2x) = 0 => sin 4x = 0`

- find the general solution to the equation, such that:

`4x = (-1)^n*sin^(-1) 0 + n*pi`

`4x = 0 + n*pi => x = (n*pi)/4`

Hence, evaluating the general solution to the given equation, yields `x = (n*pi)/4` .

If the problem requests to convert the given product into a summation or a difference, yields:

`sin 2x*cos 2x = (1/2)(sin a + sin b)`

`{(2x = (a + b)/2),(2x = (a - b)/2):} => a + b = a - b => 2b = 0 => b = 0 => a = 4x`

`sin 2x*cos 2x = (1/2)(sin (4x) + sin 0)`

Since `sin 0 = 0` yields:

`sin 2x*cos 2x = (1/2)(sin (4x))`

**Hence, converting the product into a summation, yields **`sin 2x*cos 2x = (1/2)(sin (4x)).`