Giải phương trình cos (x - 5)=sqrt3/2 với - pi  < x < pi .

$\cos (x - 5) = \frac{{\sqrt 3 }}{2} \Leftrightarrow \cos (x - 5) = \cos \frac{\pi }{6} \Leftrightarrow \left[ \begin{array}{l}x - 5 = \frac{\pi }{6} + k2\pi \\x - 5 =  - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 5 + \frac{\pi }{6} + k2\pi \\x = 5 - \frac{\pi }{6} + k2\pi \end{array} \right.$

Với điều kiện $- \pi  < x < \pi$ ta lần lượt có:

$\begin{array}{l} + )\,\, - \pi  < 5 + \frac{\pi }{6} + k2\pi  < \pi  \Leftrightarrow  - \pi  - 5 - \frac{\pi }{6} < k2\pi  < \pi  - 5 - \frac{\pi }{6}\\ \Leftrightarrow  - \frac{1}{2} - \frac{5}{{2\pi }} - \frac{1}{{12}} \approx  - 1,3 < k < \frac{1}{2} - \frac{5}{{2\pi }} - \frac{1}{{12}} \approx  - 0,3\\ \Rightarrow k =  - 1 \Rightarrow x = 5 + \frac{\pi }{6} - 2\pi  = 5 - \frac{{11\pi }}{6}.\end{array}$

$\begin{array}{l} + )\,\, - \pi  < 5 - \frac{\pi }{6} + k2\pi  < \pi  \Leftrightarrow  - \pi  - 5 + \frac{\pi }{6} < k2\pi  < \pi  - 5 + \frac{\pi }{6}\\ \Leftrightarrow  - \frac{1}{2} - \frac{5}{{2\pi }} + \frac{1}{{12}} \approx  - 1,1 < k < \frac{1}{2} - \frac{5}{{2\pi }} + \frac{1}{{12}} \approx  - 0,4\\ \Rightarrow k =  - 1 \Rightarrow x = 5 - \frac{\pi }{6} - 2\pi  = 5 - \frac{{13\pi }}{6}.\end{array}$