Giải phương trình cos ( 2x + pi /4) = 1/2, - pi < x < pi
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: \(S = \left\{ { - \frac{{23\pi }}{{24}};\,\,\frac{\pi }{{24}};\, - \frac{{7\pi }}{{24}};\,\,\frac{{17\pi }}{{24}}} \right\}\).