## Task B-3.6.

Let $f(x)=\frac{\sin 3 x}{\sin x}+\frac{\cos 3 x}{\cos x}$.

(a) Solve the equation $f(x)=2$.

(b) Determine the intervals of real numbers $x$ for which the function $f$ takes positive values.
## Solution.

First, we write the function in the form:

$$
f(x)=\frac{\sin 3 x}{\sin x}+\frac{\cos 3 x}{\cos x}=\frac{\sin 3 x \cdot \cos x+\cos 3 x \cdot \sin x}{\sin x \cdot \cos x}=\frac{\sin 4 x}{\sin x \cdot \cos x}
$$

This function is not defined for real numbers for which $\sin x \cdot \cos x=0$.

Therefore, it must be $x \neq \frac{k \pi}{2}, k \in \mathbb{Z}$. (*)

Now we have

$$
f(x)=\frac{\sin 4 x}{\sin x \cdot \cos x}=\frac{2 \sin 2 x \cdot \cos 2 x}{\sin x \cdot \cos x}=\frac{2 \sin x \cdot \cos x \cdot \cos 2 x}{\sin x \cdot \cos x}=4 \cos 2 x
$$

(a) We need to solve the equation $4 \cos 2 x=2$. This gives $\cos 2 x=\frac{1}{2}$.

The solutions are: $2 x= \pm \frac{\pi}{3}+2 k \pi$, or $x= \pm \frac{\pi}{6}+k \pi, k \in \mathbb{Z}$.

(b) The function will be positive if $4 \cos 2 x>0$, which means

$$
\begin{aligned}
& 2 x \in\left\langle-\frac{\pi}{2}+2 k \pi, \frac{\pi}{2}+2 k \pi\right\rangle, k \in \mathbb{Z} \\
& \Leftrightarrow x \in\left\langle-\frac{\pi}{4}+k \pi, \frac{\pi}{4}+k \pi\right\rangle, k \in \mathbb{Z}
\end{aligned}
$$

Due to (*), the function takes positive values for all real numbers in the interval

$$
x \in\left\langle-\frac{\pi}{4}+k \pi, k \pi\right\rangle \cup\left\langle k \pi, \frac{\pi}{4}+k \pi\right\rangle, k \in \mathbb{Z}
$$

Note: If the student solved the equation and inequality without first writing the function in the form $f(x)=4 \cos 2 x$, but instead solved each part of the problem (a) and (b) separately, then each part of the problem is worth 5 points. In part (a), deduct 1 point if the domain of the function was not considered, as the correct solution can still be obtained in that case.