<s>[INST] Find all real numbers $a,b$ such that $|a|+|b|\geq\frac{2}{\sqrt{3}}$ and $|a\sin x+b\sin{2x}|\leq 1$ for all real $x$. Find all real numbers $a,b$ such that $|a|+|b|\geqslant\frac{2}{\sqrt{3}}$ and $|a\sin(x)+b\sin(2x)|\leqslant 1$ for all real $x$.
We could write the inequality as
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\sin (x)+\frac{b}{\sqrt{a^2+b^2}}\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}}
$$
and let
$$
\cos (y)=\frac{a}{\sqrt{a^2+b^2}}
$$
and
$$
\sin (y) = \frac{b}{\sqrt{a^2+b^2}}.
$$
The inequality is equivalent to
$$
\left|\cos (y)\cos\left(\frac{\pi}{4}-x\right)-\sin (y)\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}},
$$
but the terms are different, so we cannot use the cosine addition formula.
 [/INST] If $a$ and $b$ are of the same sign then $|a|+|b|=|a+b|\geqslant\frac{2}{\sqrt3}$. It's easy to guess a value of $x$ for which inequality $|a\sin x+b\sin 2x|\leqslant 1$ becomes equality - for example, $x=\frac{\pi}{3}$. Therefore this point is a local extremum for $f(x)=a\sin x+b\sin 2x$, so $$f'(\pi/3)=a\cos\pi/3+2b\cos2\pi/3 = a/2 - b = 0,$$
i.e. $a=2b$. Using now relation $|a+b|=\frac{2}{\sqrt3}$ you can find two pairs  $(a,b)$, and both of them are suitable.
The case of $a$ and $b$ of different sign can be considered in a similar way (here $|a|+|b|=|a-b|$), and leads to another two pairs.
</s>