Phương trình $2{\cos ^2}x = 1$ có nghiệm là
A. $x = k\frac{\pi }{4}$.
B. $x = \pm \frac{\pi }{4} + k\pi $.
C. $x = k\frac{\pi }{2}$.
D. vô nghiệm.
A. $x = k\frac{\pi }{4}$.
B. $x = \pm \frac{\pi }{4} + k\pi $.
C. $x = k\frac{\pi }{2}$.
D. vô nghiệm.
Chọn B.
Ta có: $2{\cos ^2}x = 1 \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{{\sqrt 2 }}{2}\\\cos x = - \frac{{\sqrt 2 }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = - \frac{\pi }{4} + k2\pi \\x = \frac{{3\pi }}{4} + k2\pi \\x = - \frac{{3\pi }}{4} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow x = \pm \frac{\pi }{4} + k\pi ,\left( {k \in \mathbb{Z}} \right).$
Ta có: $2{\cos ^2}x = 1 \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{{\sqrt 2 }}{2}\\\cos x = - \frac{{\sqrt 2 }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = - \frac{\pi }{4} + k2\pi \\x = \frac{{3\pi }}{4} + k2\pi \\x = - \frac{{3\pi }}{4} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow x = \pm \frac{\pi }{4} + k\pi ,\left( {k \in \mathbb{Z}} \right).$