Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin \left( {\frac{{5\pi }}{4} - 6x} \right) + 15\sin \left( {\frac{\pi }{4} + 2x} \right) = 16\,\,\,\,\,\,\,\left( 1 \right)\\
- 1 \le \sin \left( {\frac{{5\pi }}{4} - 6x} \right) \le 1\\
- 1 \le \sin \left( {\frac{\pi }{4} + 2x} \right) \le 1 \Rightarrow - 15 \le 15\sin \left( {\frac{\pi }{4} + 2x} \right) \le 15\\
\Rightarrow \sin \left( {\frac{{5\pi }}{4} - 6x} \right) + 15\sin \left( {\frac{\pi }{4} + 2x} \right) \le 16\,\,\,\,\,\left( 2 \right)
\end{array}\)
Từ (1) và (2) suy ra:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sin \left( {\frac{{5\pi }}{4} - 6x} \right) = 1\\
\sin \left( {\frac{\pi }{4} + 2x} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{{5\pi }}{4} - 6x = \frac{\pi }{2} + k2\pi \\
\frac{\pi }{4} + 2x = \frac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
6x = \frac{{3\pi }}{4} + k2\pi \\
2x = \frac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \frac{\pi }{8} + \frac{{k\pi }}{3}\\
x = \frac{\pi }{8} + k\pi
\end{array} \right.\\
\Leftrightarrow x = \frac{\pi }{8} + \frac{{k\pi }}{3}\,\,\,\,\,\left( {k \in Z} \right)\\
x \in \left[ { - 2019;2019} \right]\\
\Leftrightarrow - 2019 \le \frac{\pi }{8} + \frac{{k\pi }}{3} \le 2019\\
\Leftrightarrow - 1928 \le k \le 1927
\end{array}\)
Tổng các nghiệm của pt là:
\(\begin{array}{l}
S = \left( {\frac{\pi }{8} - \frac{{1928\pi }}{3}} \right) + \left( {\frac{\pi }{8} - \frac{{1927\pi }}{3}} \right) + \left( {\frac{\pi }{8} - \frac{{1926\pi }}{3}} \right) + ..... + \left( {\frac{\pi }{8} + \frac{{1926\pi }}{3}} \right) + \left( {\frac{\pi }{8} + \frac{{1927\pi }}{3}} \right)\\
= \frac{\pi }{8}.3856 - \frac{{1928\pi }}{3}\\
= \frac{{ - 482}}{3}\pi
\end{array}\)
