Phương trình \(\sqrt 3 \sin 3x + \cos 3x =  - 1\)...

A

- Hướng dẫn giải

Phương pháp giải:

\(\begin{array}{l}a\sin x + b\cos x = c\\ \Leftrightarrow \frac{a}{{\sqrt {{a^2} + {b^2}} }}\sin x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\cos x = \frac{c}{{\sqrt {{a^2} + {b^2}} }}\,\,\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }} \in \left[ { - 1;1} \right]} \right)\\Dat\,\,\left\{ \begin{array}{l}\frac{a}{{\sqrt {{a^2} + {b^2}} }} = \cos \alpha \\\frac{b}{{\sqrt {{a^2} + {b^2}} }} = \sin \alpha \end{array} \right. \Rightarrow \sin \left( {x + \alpha } \right) = \frac{c}{{\sqrt {{a^2} + {b^2}} }}\end{array}\)

Giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\sqrt 3 \sin 3x + \cos 3x =  - 1\\ \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 3x + \frac{1}{2}\cos 3x =  - \frac{1}{2}\\ \Leftrightarrow \cos \frac{\pi }{6}\sin 3x + \sin \frac{\pi }{6}\cos 3x =  - \frac{1}{2}\\ \Leftrightarrow \sin \left( {3x + \frac{\pi }{6}} \right) =  - \frac{1}{2}\end{array}\)

Chọn: A