Nghiệm của phương trình 2cos 2x + 1 = 0 là:
A. $x = - \frac{\pi }{3} + k2\pi ;x = \frac{\pi }{3} + k2\pi $.
B. $x = - \frac{\pi }{6} + k2\pi ;x = \frac{{2\pi }}{3} + k2\pi $.
C. $x = \frac{{2\pi }}{3} + k2\pi ;x = - \frac{{2\pi }}{3} + k2\pi $.
D. $x = \frac{\pi }{3} + k\pi ;x = - \frac{\pi }{3} + k\pi $.
A. $x = - \frac{\pi }{3} + k2\pi ;x = \frac{\pi }{3} + k2\pi $.
B. $x = - \frac{\pi }{6} + k2\pi ;x = \frac{{2\pi }}{3} + k2\pi $.
C. $x = \frac{{2\pi }}{3} + k2\pi ;x = - \frac{{2\pi }}{3} + k2\pi $.
D. $x = \frac{\pi }{3} + k\pi ;x = - \frac{\pi }{3} + k\pi $.
Chọn D.
Ta có: $2\cos 2x + 1 = 0 \Leftrightarrow \cos 2x = - \frac{1}{2} \Leftrightarrow \cos 2x = \cos \frac{{2\pi }}{3} \Leftrightarrow 2x = \pm \frac{{2\pi }}{3} + k2\pi \Leftrightarrow x = \pm \frac{\pi }{3} + k\pi $.
Ta có: $2\cos 2x + 1 = 0 \Leftrightarrow \cos 2x = - \frac{1}{2} \Leftrightarrow \cos 2x = \cos \frac{{2\pi }}{3} \Leftrightarrow 2x = \pm \frac{{2\pi }}{3} + k2\pi \Leftrightarrow x = \pm \frac{\pi }{3} + k\pi $.