Which angles satisfy the following inequality:

$$
\cos 2 x-4 \cos \frac{\pi}{4} \cos \frac{5 \pi}{12} \cos x+\cos \frac{5 \pi}{6}+1>0
$$
Let's input the known cosine values on the left side:

$$
\cos (5 \pi / 6)=\cos 150^{\circ}=-\sqrt{3} / 2, \quad \cos \pi / 4=1 / \sqrt{2}
$$

$\cos (5 \pi / 12)=\cos \left(45^{\circ}+30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}=\sqrt{2}(\sqrt{3}-1) / 4$, so the coefficient of the second term is: $-(\sqrt{3}-1)$. Considering the identity $\cos 2 x=2 \cos ^{2} x-1$, the inequality becomes:

$$
2 \cos ^{2} x-(\sqrt{3}-1) \cos x-\sqrt{3} / 2>0
$$

and by factoring the left side as a quadratic expression in $\cos x$ and simplifying by the positive 2, we get:

$$
\left(\cos x+\frac{1}{2}\right)\left(\cos x-\frac{\sqrt{3}}{2}\right)>0
$$

The points $-1 / 2$ and $\sqrt{3} / 2$, where the product can change sign, both fall within the closed interval of values that $\cos x$ can take, i.e., [-1, 1]. Thus, as we move through this interval, both factors change sign. (1) holds if both factors have the same sign, i.e., if the smaller one is positive, or if the larger one is negative, which means:

$$
\begin{array}{cc}
\text { or } & \cos x-\frac{\sqrt{3}}{2}>0, \quad \text { i.e., } \quad \cos x>\frac{\sqrt{3}}{2} \\
\text { or } & \cos x+\frac{1}{2}<0, \quad \text { i.e., } \quad \cos x<-\frac{1}{2}
\end{array}
$$

We can also arrive at this from the form:

$$
\begin{gathered}
-2 \cos \frac{\pi}{4} \cos \frac{5 \pi}{12}=-\cos \left(\frac{\pi}{4}+\frac{5 \pi}{12}\right)-\cos \left(\frac{\pi}{4}-\frac{5 \pi}{12}\right)=-\left(\cos \frac{2 \pi}{3}+\cos \frac{\pi}{6}\right) \\
\frac{1}{2} \cos \frac{5 \pi}{6}=\cos \frac{5 \pi}{6} \cos \frac{\pi}{3}=\left(-\cos \frac{\pi}{6}\right)\left(-\cos \frac{2 \pi}{3}\right)=\cos \frac{2 \pi}{3} \cos \frac{\pi}{6}
\end{gathered}
$$

Here, we are dealing with the sum and product of the same two cosine values with opposite signs, so (2) can also be written as:

$$
\left(\cos x-\cos \frac{2 \pi}{3}\right)\left(\cos x-\cos \frac{\pi}{6}\right)>0
$$

From here, we can proceed according to the above solution.