A1. For the function $f$, it is $3 f(x)+f(-x)=4 \sin x \cos x$ for any real number $x$. Find the correct expression for the function $f$.
(A) $\sin x$
(B) $\cos x$
(C) $\cos x \sin x$
(D) $\sin 2 x$
(E) $\cos 2 x$
A1. From the equality, we express $f(x)=\frac{4}{3} \sin x \cos x-\frac{1}{3} f(-x)$. If we substitute $-x$ for $x$ in the equation and consider that the sine function is odd and the cosine function is even, we get $f(-x)=$ $\frac{4}{3} \sin (-x) \cos (-x)-\frac{1}{3} f(x)=-\frac{4}{3} \sin x \cos x-\frac{1}{3} f(x)$. When we substitute the latter into the first equation, we get $f(x)=\frac{4}{3} \sin x \cos x-\frac{1}{3}\left(-\frac{4}{3} \sin x \cos x-\frac{1}{3} f(x)\right)=\left(\frac{4}{3}+\frac{4}{9}\right) \sin x \cos x+\frac{1}{9} f(x)$. We rearrange the equation to $\frac{8}{9} f(x)=\frac{16}{9} \sin x \cos x$. From this, it follows that $f(x)=2 \sin x \cos x=\sin 2 x$. The correct answer is $D$.