Giải phương trình \(3\cos x + 2\sqrt 3 \sin x = \frac{9}{2}\).

Lời giải:

Chia 2 vế PT đề bài ra cho \(\sqrt {21} \), ta được: \(\frac{{\sqrt {21} }}{7}\cos x + \frac{{2\sqrt 7 }}{7}{\mathop{\rm sinx}\nolimits} = \frac{{3\sqrt {21} }}{{14}}\) (*).

Đặt \(\sin \alpha = \frac{{\sqrt {21} }}{7};\cos \alpha = \frac{{2\sqrt 7 }}{7}\)

(*) \( \Leftrightarrow \sin \alpha .cos\alpha + cos\alpha .sin\alpha = \frac{{3\sqrt {21} }}{{14}}\)

\( \Leftrightarrow \sin \left( {x + \alpha } \right) = \frac{{3\sqrt {21} }}{{14}}\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \alpha = \arcsin \frac{{3\sqrt {21} }}{{14}} + k2\pi }\\{x + \alpha = \pi - \arcsin \frac{{3\sqrt {21} }}{{14}} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \arcsin \frac{{3\sqrt {21} }}{{14}} - \alpha + k2\pi }\\{x = \pi - \alpha - \arcsin \frac{{3\sqrt {21} }}{{14}} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).