Nghiệm của phương trình \(2{\cos ^2}\left( {2x + \frac{\pi }{3}} \right) + 3\cos \left( {2x + \frac{\pi }{3}} \right) - 5 = 0\) trong khoảng \(\left( { - \frac{{3\pi }}{2};\frac{{3...

* Hướng dẫn giải

\(\begin{array}{l} 2{\cos ^2}\left( {2x + \frac{\pi }{3}} \right) + 3\cos \left( {2x + \frac{\pi }{3}} \right) - 5 = 0 \Leftrightarrow \left[ \begin{array}{l} \cos \left( {2x + \frac{\pi }{3}} \right) = 1\\ \cos \left( {2x + \frac{\pi }{3}} \right) = - \frac{5}{2}\left( {Loai} \right) \end{array} \right.\\ \cos \left( {2x + \frac{\pi }{3}} \right) = 1 \Leftrightarrow 2x + \frac{\pi }{3} = k2\pi & \Leftrightarrow x = - \frac{\pi }{6} + k\pi \,\,\left( {k \in Z} \right) \end{array}\)

Theo đề ra \(- \frac{{3\pi }}{2} < x = - \frac{\pi }{6} + k\pi < \frac{{3\pi }}{2} \Leftrightarrow - \frac{4}{3} < k < \frac{5}{3} \Rightarrow \left[ \begin{array}{l} k = - 1\\ k = 0\\ k = 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = - \frac{{7\pi }}{6}\\ x = - \frac{\pi }{6}\\ x = \frac{{5\pi }}{6} \end{array} \right.\)