Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
1.{\mathop{\rm cosx}\nolimits} = \frac{{\sqrt 3 }}{2}\\
= > x = \pm \frac{\pi }{6} + k2\pi \\
2.2{\sin ^2}x - 3\sin x - 2 = 0\\
= > \left[ \begin{array}{l}
\sin x = - \frac{1}{2} = > \left[ \begin{array}{l}
x = \frac{{ - \pi }}{6} + k2\pi \\
x = \frac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\sin x = 2 = > v{n_o}
\end{array} \right.
\end{array}\]
\[\begin{array}{l}
3.\sin 2x + \cos 2x = 1\\
\Leftrightarrow \sin (2x + \frac{\pi }{4}) = \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\
2x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi
\end{array} \right. = > \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k\pi
\end{array} \right.
\end{array}\]
\[\begin{array}{l}
4.cos7x.cos5x - \sqrt 3 \sin 2x = 1 - \sin 7x.\sin 5x\\
\Leftrightarrow (cos7x.cos5x + \sin 7x.\sin 5x) - \sqrt 3 \sin 2x = 1\\
\Leftrightarrow \cos (7x - 5x) - \sqrt 3 \sin 2x = 1\\
\Leftrightarrow \cos 2x - \sqrt 3 \sin 2x = 1\\
\Leftrightarrow \cos (2x + \frac{\pi }{3}) = 1\\
\Leftrightarrow 2x + \frac{\pi }{3} = k2\pi = > x = - \frac{\pi }{6} + k\pi
\end{array}\]
