Đáp án:
\(x \in \left\{ {\frac{{ - \pi }}{2},0,\frac{\pi }{2},\pi ,\frac{{3\pi }}{2},2\pi ,\frac{{5\pi }}{2},3\pi ,\frac{{7\pi }}{2}} \right\}\)
Giải thích các bước giải:
cos8x+cos4x-2=0
<-> 2cos²4x-1+cos4x-2=0
<-> 2cos²4x+cos4x-3=0
<-> cos4x=$\frac{-3}{2}$ (Vô nghiệm ) hoặc cos4x=1
-> \(\begin{array}{l}
4x = k2\pi \leftrightarrow x = \frac{{k\pi }}{2}(k \in Z)\\
x \in \left[ {\frac{{ - \pi }}{2},\frac{{11\pi }}{3}} \right]\\
\to \frac{{ - \pi }}{2} \le \frac{{k\pi }}{2} \le \frac{{11\pi }}{3} \leftrightarrow - 1 \le k \le \frac{{22}}{3},k \in Z \to k \in \left[ { - 1,7} \right]\\
\to x \in \left\{ {\frac{{ - \pi }}{2},0,\frac{\pi }{2},\pi ,\frac{{3\pi }}{2},2\pi ,\frac{{5\pi }}{2},3\pi ,\frac{{7\pi }}{2}} \right\}
\end{array}\)
