Lời giải

Ta có \(\cos \left( {2x + \frac{\pi }{4}} \right) = \frac{1}{2}\).

\[ \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\2x + \frac{\pi }{4} = - \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{{12}} + k2\pi \\2x = - \frac{{7\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{24}} + k\pi \\x = - \frac{{7\pi }}{{24}} + k\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\].

• Với –π < x < π, ta có: \( - \pi < \frac{\pi }{{24}} + k\pi < \pi \).

\( \Leftrightarrow - \frac{{25\pi }}{{24}} < k\pi < \frac{{23\pi }}{{24}}\)

\( \Leftrightarrow - \frac{{25}}{{24}} < k < \frac{{23}}{{24}}\).

Mà k ∈ ℤ.

Suy ra k ∈ {–1; 0}.

Khi đó \(x = - \frac{{23\pi }}{{24}};\,x = \frac{\pi }{{24}}\).

• Với –π < x < π, ta có: \( - \pi < - \frac{{7\pi }}{{24}} + k\pi < \pi \).

\( \Leftrightarrow - \frac{{17\pi }}{{24}} < k\pi < \frac{{31\pi }}{{24}}\)

\( \Leftrightarrow - \frac{{17}}{{24}} < k < \frac{{31}}{{24}}\).

Mà k ∈ ℤ.

Suy ra k ∈ {0; 1}.

Khi đó \(x = - \frac{{7\pi }}{{24}};\,\,x = \frac{{17\pi }}{{24}}\).

Vậy tập nghiệm của phương trình đã cho là: \(S = \left\{ { - \frac{{23\pi }}{{24}};\,\,\frac{\pi }{{24}};\, - \frac{{7\pi }}{{24}};\,\,\frac{{17\pi }}{{24}}} \right\}\).