Đáp án:
$\left[\begin{array}{l}x = -2\pi\\x = 0\\x = 2\pi\end{array}\right.$
Giải thích các bước giải:
$\begin{array}{l}\cos^2x + 2\cos x - 3 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = 1\\\cos x = - 3\quad (loại)\end{array}\right.\\ \Leftrightarrow x = k2\pi\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ -2\pi \leq x \leq 2\pi\\ \Leftrightarrow -2\pi \leq k2\pi \leq 2\pi\\ \Leftrightarrow -1 \leq k \leq 1\\ Do\,\,k \in \Bbb Z\\ nên\,\,k = \left\{-1;0;1\right\}\\ Ta\,\,được:\\ \quad \left[\begin{array}{l}x = -2\pi\\x = 0\\x = 2\pi\end{array}\right. \end{array}$
