Obviously $a>1$ otherwise, putting $x=\frac{\pi}{2}$ gives $|\cos \frac{a \pi}{2} | > 1+\sin \frac{a \pi}{2} $ which is contradiction since LHS is $\le 1$ and $RHS >1$. Suppose $a>1$ then we take , $ax=u, a=\frac{1}{m}$ which gives $|\cos mu|+|\cos u|>\sin u+\sin mu \forall u \in R$. Clearly $m \le 1$ which is contradiction.

To add to the previous answer, $a=0$ is an obvious solution ..... there seems to be no other solution

lebleu wrote: To add to the previous answer, $a=0$ is an obvious solution ..... there seems to be no other solution In the problem statement it is said that "a" is positive.

Firstly, if $a$ is irrational, then we can take $x=\frac{\pi}{2}+2\pi k$. Taking $k$ such that $ax$ mod $2 \pi$ is close to $\frac{\pi}{2}$, we win. If $a$ is rational, say $a=\frac{p}{q}$, then WLOG assume $p$ is odd(otherwise change $a$ to $\frac{1}{a}$). Putting $x=\frac{q \pi}{2}$ yields a contradiction.