Solve the following equation:

$$
7 \cos ^{3} z-6 \cos z=3 \cos 3 z
$$
The right side can be transformed based on known identities as follows:

$$
\begin{aligned}
3 \cos 3 z & =3 \cos (z+2 z)=3 \cos z \cos 2 z-3 \sin z \sin 2 z= \\
& =3 \cos z\left(\cos ^{2} z-\sin ^{2} z\right)-6 \sin ^{2} z \cos z= \\
& =3 \cos z\left(\cos ^{2} z-3 \sin ^{2} z\right)=3 \cos z\left(4 \cos ^{2} z-3\right)= \\
& =12 \cos ^{3} z-9 \cos z
\end{aligned}
$$

Now we reduce to 0 and factor the polynomial of the equation:

$$
\begin{aligned}
0 & =5 \cos ^{3} z-3 \cos z=5 \cos z\left(\cos ^{2} z-0.6\right)= \\
& =5 \cos z(\cos z-\sqrt{0.6})(\cos z+\sqrt{0.6})
\end{aligned}
$$

From here, the solutions are:

$$
\begin{array}{ll}
\cos z=0 \text { gives: } & z=90^{\circ}+k \cdot 180^{\circ} \\
\cos z-\sqrt{0.6}=0 \text { gives: } & z=k \cdot 360^{\circ} \pm 39^{\circ} 14' \\
\cos z+\sqrt{0.6}=0 \text { gives: } & z=k \cdot 360^{\circ} \pm 140^{\circ} 46'
\end{array}
$$

where $k$ is an integer. - We performed the same transformations, no other solutions exist.

Note. The last two results of the solution can also be written as:

$$
z=39^{\circ} 14'+k \cdot 180^{\circ}, \quad z=140^{\circ} 46'+k \cdot 180^{\circ}
$$