### Instruction: If $\sin x = 3 \cos x,$ then what is $\sin x \cos x$?### Response: To find $\sin x \cos x,$ I need to use the fact that $\sin x = 3 \cos x.$
I can rearrange this equation to get $\cos x = \frac{\sin x}{3}.$
Then I can substitute this into $\sin x \cos x$ and get $\sin x \cos x = \sin x \cdot \frac{\sin x}{3}.$
This simplifies to $\sin x \cos x = \frac{\sin^2 x}{3}.$
Now, I need to find $\sin^2 x.$
I can use the identity $\sin^2 x + \cos^2 x = 1$ and substitute $\cos x = \frac{\sin x}{3}$ again.
This gives me $\sin^2 x + \left(\frac{\sin x}{3}\right)^2 = 1.$
Multiplying everything by $9,$ I get $9 \sin^2 x + \sin^2 x = 9.$
Combining like terms, I get $10 \sin^2 x = 9.$
Dividing by $10,$ I get $\sin^2 x = \frac{9}{10}.$
Plugging this into $\sin x \cos x = \frac{\sin^2 x}{3},$ I get $\sin x \cos x = \frac{9}{10} \cdot \frac{1}{3}.$
Multiplying, I get $\sin x \cos x = \frac{3}{10}.$