Alternatively, You can use Eulers relation which is often handy when considering products of trig functions. Eulers formula states that the complex exponential e^ix = cox(x) + i*sin(x). This formula can be solved for sin and cos to get the following identities:

cos(x) = ( e^ix + e^-ix)/2

sin(x) = (e^ix - e^-ix)/2i

So, the solution becomes:

6*cosx*sinx = 6*( e^ix + e^-ix)*(e^ix - e^-ix)/4i

= 3(e^ix * e^ix + e^ix * e^-ix - e^ix * e^-ix - e^-ix * e^-ix) / 2i

= 3(e^i2x - e^-i2x)/2i

= 3 sin(2x)

We have to express 6*(sin x)*(cos x) as a single sine or cosine function.

Now we know that sin 2x = 2*(sin x)*(cos x).

We use this relation here.

6 sin x cos x

=> 3*2*(sin x)*(cos x)

=> 3*sin 2x

**Therefore 6*(sin x)*(cos x) = 3*sin 2x.**

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