Cho \(x \in \left( {0;\dfrac{\pi }{2}} \right)\). Biết \(\log \sin x + \log \cos x = - 1\) và \(\log \left( {\sin x + \cos x} \right) = \dfrac{1}{2}\left( {\log n - 1} \right)\)....

* Hướng dẫn giải

Ta có: \(\log \sin x + \log \cos x =  - 1\) \( \Leftrightarrow \log \left( {\sin x\cos x} \right) =  - 1\) \( \Leftrightarrow \sin x\cos x = \dfrac{1}{{10}}\)

\(\log \left( {\sin x + \cos x} \right) = \dfrac{1}{2}\left( {\log n - 1} \right)\)\( \Leftrightarrow 2\log \left( {\sin x + \cos x} \right) = \log n - 1\) \( \Leftrightarrow \log {\left( {\sin x + \cos x} \right)^2} = \log \dfrac{n}{{10}}\)\( \Leftrightarrow \log \left( {1 + 2\sin x\cos x} \right) = \log \dfrac{n}{{10}}\) \( \Leftrightarrow 1 + 2\sin x\cos x = \dfrac{n}{{10}}\)\( \Rightarrow 1 + 2.\dfrac{1}{{10}} = \dfrac{n}{{10}} \Leftrightarrow n = 12\).

Chọn B