Số nghiệm của phương trình sau \(\sin 5x + \sqrt 3 \cos 5x = 2\sin 7x\) trên khoảng \(\left( {0;\dfrac{\pi }{2}} \right)\)là?

* Hướng dẫn giải

Ta có

\(\begin{array}{l}\,\,\,\,\,\,\sin 5x + \sqrt 3 \cos 5x = 2\sin 7x\\ \Leftrightarrow \dfrac{1}{2}\sin 5x + \dfrac{{\sqrt 3 }}{2}\cos 5x = \sin 7x\\ \Leftrightarrow \sin 5x\cos \dfrac{\pi }{3} + \cos 5x\sin \dfrac{\pi }{3} = \sin 7x\\ \Leftrightarrow \sin \left( {5x + \dfrac{\pi }{3}} \right) = \sin 7x\\ \Leftrightarrow \left[ \begin{array}{l}7x = 5x + \dfrac{\pi }{3} + k2\pi \\7x =  - 5x + \dfrac{{2\pi }}{3} + m2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{m\pi }}{6}\end{array} \right.\,\,\left( {k,\,\,m \in \mathbb{Z}} \right)\end{array}\)

Mà \(x \in \left( {0;\dfrac{\pi }{2}} \right)\)\( \Rightarrow \left[ \begin{array}{l}0 < \dfrac{\pi }{6} + k\pi  < \dfrac{\pi }{2}\\0 < \dfrac{\pi }{{18}} + \dfrac{{m\pi }}{6} < \dfrac{\pi }{2}\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l} - \dfrac{1}{6} < k < \dfrac{1}{3}\\ - \dfrac{1}{3} < m < \dfrac{8}{3}\end{array} \right.\)\( \Rightarrow \left[ \begin{array}{l}k = 0\\m \in \left\{ {0;1;2} \right\}\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6}\\x \in \left\{ {\dfrac{\pi }{{18}};\dfrac{{2\pi }}{9};\dfrac{{7\pi }}{{18}}} \right\}\end{array} \right.\)

Vậy phương trình đã cho có 4 nghiệm thỏa mãn.

Chọn A.