Câu hỏi 11

\(\begin{array}{l}\,\,\,\,\,\,{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)^2} + \sqrt 3 \cos x = 2\\ \Leftrightarrow 1 + 2\sin \dfrac{x}{2}.\cos \dfrac{x}{2} + \sqrt 3 \cos x = 2\\ \Leftrightarrow \sin x + \sqrt 3 \cos x = 1\\ \Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \dfrac{1}{2}\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{6}} \right).\sin x + \cos \left( {\dfrac{\pi }{6}} \right).\cos x = \dfrac{1}{2}\\ \Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{3}\\ \Leftrightarrow \left[ \begin{array}{l}x - \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\x - \dfrac{\pi }{6} =  - \dfrac{\pi }{3} + l2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\x =  - \dfrac{\pi }{6} + l2\pi \end{array} \right.\,\,\,\left( {k,\,\,l \in \mathbb{Z}} \right)\end{array}\)

Mà \(x \in \left[ {0;\pi } \right]\), do đó \(\left\{ \begin{array}{l}0 \le \dfrac{\pi }{2} + k2\pi  \le \pi \\0 \le  - \dfrac{\pi }{6} + l2\pi  \le \pi \end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l} - \dfrac{1}{4} \le k \le \dfrac{1}{4}\\\dfrac{1}{{12}} \le l \le \dfrac{7}{{12}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = 0\\l \in \emptyset \end{array} \right..\)

Vậy phương trình đã cho có 1 nghiệm \(x = \dfrac{\pi }{2}\) thỏa mãn yêu cầu bài toán.

Chọn B.