5. For given natural numbers $k_{0}<k_{1}<k_{2}$, determine the smallest number of roots on the interval $[0 ; 2 \pi)$ that the equation of the form

$$
\sin k_{0} x+A_{1} \sin k_{1} x+A_{2} \sin k_{2} x=0
$$

where $A_{1}, A_{2} \in \mathbb{R}$, can have.
5. Let $N(F)$ denote the number of zeros of the function $F$ on the half-interval $[0 ; 2 \pi)$, i.e., the number of values of the argument $x \in [0 ; 2 \pi)$ for which $F(x)=0$. Then, for $A_{1}=A_{2}=0$, we have $N\left(\sin k_{0} x\right)=2 k_{0}$. Indeed, the zeros are the numbers $x_{n}=\frac{\pi n}{k_{0}}$, where $n=0,1, \ldots, 2 k_{0}-1$.

Fix arbitrary numbers $A_{1}$ and $A_{2}$ and prove that the number of zeros of the function

$$
F(x)=\sin k_{0} x+A_{1} \sin k_{1} x+A_{2} \sin k_{2} x
$$

on the half-interval $\left[0 ; 2 \pi\right)$ is not less than $2 k_{0}$. Denote

$$
f_{m}(x)=\left\{\begin{aligned}
\sin x, & \text { if } m \text { is divisible by } 4, \\
-\cos x, & \text { if } m-1 \text { is divisible by } 4, \\
-\sin x, & \text { if } m-2 \text { is divisible by } 4, \\
\cos x, & \text { if } m-3 \text { is divisible by } 4 .
\end{aligned}\right.
$$

Clearly, $f_{m+1}^{\prime}(x)=f_{m}(x)$. Now define the sequence of functions

$$
F_{m}(x)=f_{m}\left(k_{0} x\right)+A_{1}\left(\frac{k_{0}}{k_{1}}\right)^{m} f_{m}\left(k_{1} x\right)+A_{2}\left(\frac{k_{0}}{k_{2}}\right)^{m} f_{m}\left(k_{2} x\right)
$$

$m=0,1, \ldots$ Then $F_{0}=F$ and $F_{m+1}^{\prime}=k_{0} F_{m}$. It is clear that the period of each of the functions $F_{m}$ is $2 \pi$.

Lemma. Let $f$ be a differentiable function with period $2 \pi$. Then the number of zeros of the function $f$ on the half-interval $[0 ; 2 \pi)$ does not exceed the number of zeros of its derivative on the same half-interval.

Proof. We use Rolle's theorem: between any two zeros of a differentiable function, there is at least one zero of its derivative (see the comment to the problem). Let $x_{1}, x_{2}, \ldots, x_{N}$ be the zeros of the function on the specified half-interval. By Rolle's theorem, on each of the intervals $\left(x_{1} ; x_{2}\right),\left(x_{2} ; x_{3}\right), \ldots,\left(x_{N-1} ; x_{N}\right),\left(x_{N} ; x_{1}+2 \pi\right)$, there is at least one zero of the derivative. However, the zero of the derivative on the last interval (denote it by $y$) may lie outside the half-interval $[0 ; 2 \pi)$. In this case, consider $y-2 \pi$—this is also a zero of the derivative, since the derivative of a periodic function is periodic. The lemma is proved.

From the lemma, it follows that $N\left(F_{m}\right) \geqslant N\left(F_{m+1}\right)$. Therefore, it is sufficient to prove that $N\left(F_{M}\right) \geqslant 2 k_{0}$ for a sufficiently large number $M$.

Since $\frac{k_{0}}{k_{1}}1-\varepsilon>0$ for even $n$ and $F_{M}\left(\frac{\pi n}{k_{0}}\right)<-1+$ $+\varepsilon<0$ for odd $n$. Therefore, the continuous function $F_{M}$ must have a zero between any two adjacent points $x_{n}=\frac{\pi n}{k_{0}}$, where $n=0,1, \ldots, 2 k_{0}$. Hence, $N\left(F_{M}\right) \geqslant 2 k_{0}$.

Comments. $1^{\circ}$. A periodic function can be thought of as a function on a circle. Then the statement of the lemma can be reformulated as follows: the number of zeros of a function on a circle does not exceed the number of zeros of its derivative, and the proof becomes more transparent.

$2^{\circ}$. The idea of the proof of Rolle's theorem: consider the extrema (maximum and minimum) of the function on the interval connecting the points where the function is zero. If one of the extrema is attained inside the interval, then the derivative is zero at that point. If both extrema are attained at the endpoints of the interval, then the function is identically zero on that interval.

$3^{\circ}$. In general, a function may have infinitely many zeros on the half-interval. In this case, the lemma should be understood as follows: if $f$ has infinitely many zeros on the half-interval, then $f^{\prime}$ also has infinitely many zeros on that half-interval. Verify that the proof remains valid in this case.