Problem 11.3. Let $k_{1}$ be the smallest natural number that is a root of the equation

$$
\sin k^{\circ}=\sin 334 k^{\circ}
$$

(a) (2 points) Find $k_{1}$.

(b) (2 points) Find the smallest root of this equation that is a natural number greater than $k_{1}$.
# Answer:

(a) (2 points) 36.

(b) (2 points) 40.

Solution.

$$
0=\sin 334 k^{\circ}-\sin k^{\circ}=2 \sin \frac{333 k^{\circ}}{2} \cos \frac{335 k^{\circ}}{2}
$$

- Let $\sin \frac{333 k^{\circ}}{2}=0$. This is equivalent to $\frac{333 k}{2}=180 \mathrm{~m}$ for some integer $m$, that is, $333 k=360 m$ and $37 k=40 m$. Since $k$ is a natural number, $k$ must be divisible by 40, that is, $k \geqslant 40$. It is also clear that $k=40$ is a root (corresponding to $m=37$).
- Let $\cos \frac{335 k^{\circ}}{2}=0$. This is equivalent to $\frac{335 k}{2}=90+180 n$ for some integer $n$, that is, $335 k=360 n+180$ and $67 k=36(2 n+1)$. Since $k$ is a natural number, $k$ must be divisible by 36, that is, $k \geqslant 36$. It is also clear that $k=36$ is a root (corresponding to $m=33$). The next largest root in this series of solutions is greater than 40.

Thus, the smallest natural root of the original equation is 36, and the next largest is 40.