3. Find the set of values and the smallest period of the function $y=\cos ^{2} 3 x+\sin 6 x$.
Answer. The set of values $\left[\frac{1}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}+\frac{\sqrt{5}}{2}\right]$, the smallest period $\pi / 3$. Hint. Transform the function $y$ to the form $y=\frac{1+\cos 6 x}{2}+\sin 6 x=\frac{1}{2}+\frac{\sqrt{5}}{2} \sin (6 x+\alpha)$, where $\alpha-$ is an auxiliary angle ( $\sin \alpha=\sqrt{1 / 5}, \cos \alpha=\sqrt{4 / 5}$ ). Thus, the graph of this function is obtained from the graph of $y=\sin 6 x$ by stretching along the $y$-axis by a factor of $\frac{\sqrt{5}}{2}$, then shifting along the $x$-axis to the left by $\frac{\alpha}{6}$, and finally shifting upwards along the $y$-axis by $\frac{1}{2}$. Therefore, the answer to the question follows from the fact that the set of values of the function $y=\sin 6 x$ is $[-1 ; 1]$, and the smallest period is $\frac{2 \pi}{6}$; hence, the set of values of the given function is $\left[\frac{1}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}+\frac{\sqrt{5}}{2}\right]$, and the period is the same as that of the function $y=\sin 6 x$, i.e., $\frac{\pi}{3}$.