7. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, etc.)
Solution: Let $t=\cos \varphi$. The equation will take the form: $32 \cos ^{5} \varphi-40 \cos ^{3} \varphi+10 \cos \varphi=\sqrt{3}$. Transform the left side:

\[
\begin{aligned}
& 2 \cos \varphi\left(16 \cos ^{4} \varphi-20 \cos ^{2} \varphi+5\right)=[\text{reduction formulas}]= \\
& =2 \cos \varphi\left(4(1+\cos 2 \varphi)^{2}-10-10 \cos 2 \varphi+5\right)= \\
& =2 \cos \varphi\left(4 \cos ^{2} 2 \varphi-2 \cos 2 \varphi-1\right)=2 \cos \varphi(2(1+\cos 4 \varphi)-2 \cos 2 \varphi-1) \\
& =2 \cos \varphi(-4 \sin 3 \varphi \sin \varphi+1)=-4 \sin 3 \varphi \sin 2 \varphi+2 \cos \varphi= \\
& =-2(\cos \varphi-\cos 5 \varphi)+2 \cos \varphi=2 \cos 5 \varphi
\end{aligned}
\]

Finally, $\cos 5 \varphi=\sqrt{3} / 2$. Hence, $\varphi= \pm \frac{\pi}{30}+\frac{2 \pi n}{5}, n \in \mathbb{Z}$. Since the original equation has exactly five real roots (by condition), to present them, it is sufficient to take any five values of $\varphi$ whose cosines are distinct. For example, $\varphi \in\left\{6^{\circ}\right.$, $\left.78^{\circ}, 150^{\circ}, 222^{\circ}, 294^{\circ}\right\}$.

Answer: The other four roots are of the form $t=\cos \varphi$, where $\varphi \in\left\{78^{\circ}, 150^{\circ}, 222^{\circ}, 294^{\circ}\right\}$.