Pt đã cho tương đương với:

$2 \sin 17x + (\sin 5x + \sqrt{3} \cos 5x) = 0$

Ta có:

$\sin 5x + \sqrt{3} \cos 5x = 2 \left( \frac{1}{2} \sin 5x + \frac{\sqrt{3}}{2} \cos 5x \right)$

$= 2 \left( \sin 5x \cos \frac{\pi}{3} + \cos 5x \sin \frac{\pi}{3} \right)$

$= 2 \sin \left( 5x + \frac{\pi}{3} \right)$

$2 \sin 17x + 2 \sin \left( 5x + \frac{\pi}{3} \right) = 0$

$\sin 17x + \sin \left( 5x + \frac{\pi}{3} \right) = 0$

 $\sin A + \sin B = 2 \sin\frac{A+B}{2} \cos\frac{A-B}{2}$:

$2 \sin \left( \frac{17x + 5x + \frac{\pi}{3}}{2} \right) \cos \left( \frac{17x - \left(5x + \frac{\pi}{3}\right)}{2} \right) = 0$

$2 \sin \left( \frac{22x + \frac{\pi}{3}}{2} \right) \cos \left( \frac{12x - \frac{\pi}{3}}{2} \right) = 0$

$\sin \left( 11x + \frac{\pi}{6} \right) \cos \left( 6x - \frac{\pi}{6} \right) = 0$

$\left[ \begin{array}{l} \sin \left( 11x + \frac{\pi}{6} \right) = 0 \\ \cos \left( 6x - \frac{\pi}{6} \right) = 0 \end{array} \right.$

TH 1

$\sin \left( 11x + \frac{\pi}{6} \right) = 0$

$11x + \frac{\pi}{6} = k\pi$

$11x = -\frac{\pi}{6} + k\pi$

$x = -\frac{\pi}{66} + \frac{k\pi}{11} \quad (k \in \mathbb{Z})$

TH2

$\cos \left( 6x - \frac{\pi}{6} \right) = 0$

$6x - \frac{\pi}{6} = \frac{\pi}{2} + k\pi$

$6x = \frac{\pi}{2} + \frac{\pi}{6} + k\pi$

$6x = \frac{4\pi}{6} + k\pi$

$6x = \frac{2\pi}{3} + k\pi$

$x = \frac{\pi}{9} + \frac{k\pi}{6} \quad (k \in \mathbb{Z})$

Vậy, họ nghiệm của phương trình là:

$$ \left[ \begin{array}{l} x = -\frac{\pi}{66} + \frac{k\pi}{11} \\ x = \frac{\pi}{9} + \frac{k\pi}{6} \end{array} \right. \quad (k \in \mathbb{Z}) $$right. \quad (k \in \mathbb{Z}) $$