## Task B-3.6.

Determine the total length of all intervals of real numbers that are solutions to the system of inequalities

$$
12^{\sin (2 x)} \cdot 14^{\frac{1}{\lg x+\operatorname{ctg} x}} \geqslant \sqrt[4]{2016}, \quad|x| \leqslant 2016 \pi
$$
## Solution.

The solution to the inequality $|x| \leqslant 2016 \pi$ is the interval $[-2016 \pi, 2016 \pi]$.

Simplify the remaining inequality.

$$
\begin{array}{ll}
12^{2 \sin x \cos x} \cdot 14^{\frac{\sin x \cos x}{\sin ^{2} x+\cos 2} x} \geqslant 2016^{\frac{1}{4}} & \\
144^{\sin x \cos x} \cdot 14^{\sin x \cos x} \geqslant 2016^{\frac{1}{4}} & 2 \text { points } \\
(144 \cdot 14)^{\sin x \cos x} \geqslant 2016^{\frac{1}{4}} & 1 \text { point } \\
2016^{\sin x \cos x} \geqslant 2016^{\frac{1}{4}} & \\
\sin x \cos x \geqslant \frac{1}{4} & 1 \text { point } \\
2 \sin x \cos x \geqslant \frac{1}{2} & 1 \text { point } \\
\sin (2 x) \geqslant \frac{1}{2} & \\
\frac{\pi}{6}+2 k \pi \leqslant 2 x \leqslant \frac{5 \pi}{6}+2 k \pi, k \in \mathbb{Z} & \\
\frac{\pi}{12}+k \pi \leqslant x \leqslant \frac{5 \pi}{12}+k \pi, k \in \mathbb{Z} &
\end{array}
$$

Thus, the solution to the first inequality is the union of all intervals

$$
\left[\frac{\pi}{12}+k \pi, \frac{5 \pi}{12}+k \pi\right], k \in \mathbb{Z}
$$

All obtained intervals have a length of

$$
\frac{5 \pi}{12}-\frac{\pi}{12}=\frac{\pi}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-13.jpg?height=63&width=1551&top_left_y=1573&top_left_x=184)
is $4032 \frac{\pi}{3}=1344 \pi$.