Phương trình tương đương với:sinx+12(sin3x+sinx)−√3cos3x=2(cos5x+sin3x)⇔32sinx−2sin3x+12sin3x−√3cos3x=2cos5x $\Leftrightarrow
\frac{1}{2}\sin 3x +
\frac{1}{2}\sin 3x -\sqrt{3}
\cos 3x=2\cos 5x \Leftrightarrow
\frac{1}{2}\sin 3x -\frac{\sqrt{3}}{2} \cos 3x=\cos 5x \Leftrightarrow \sin\Big(3x-\frac{\pi}{3}\Big)=\sin\Big(\frac{\pi}{2}- 5x\Big)\Leftrightarrow \left[ \begin{array}{l} 3x-\frac{\pi}{3}=\frac{\pi}{2}-5x+2k\pi\\
\frac{\pi}{3}-3x=5x+\frac{\pi}{2}+2k\pi \end{array} \right. (k\in\mathbb{Z})$$\Leftrightarrow \left[ \begin{array}{l} x=\frac{5\pi}{48}+\frac{k}{4}\pi\\ x=\frac{-\pi}{48}-\frac{k}{4}\pi \end{array} \right. (k\in\mathbb{Z})$
Phương trình tương đương với:
sinx+12(sin3x+sinx)−√3cos3x=2(cos5x+sin3x)⇔32sinx−2sin3x+12sin3x−√3cos3x=2cos5x ⇔12sin3x+12sin3x−√3cos3x=2cos5x ⇔12sin3x−√32cos3x=cos5x ⇔sin(3x−π3)=sin(π2−5x) $\Leftrightarrow \left[ \begin{array}{l} 3x-\frac{\pi}{3}=\frac{\pi}{2}-5x+2k\pi\\
3x- \frac{\pi}{3}=5x+\frac{\pi}{2}+2k\pi \end{array} \right. (k\in\mathbb{Z})$$\Leftrightarrow \left[ \begin{array}{l} x=\frac{5\pi}{48}+\frac{k}{4}\pi\\ x=\frac{-
5\pi}{
12}-k\pi \end{array} \right. (k\in\mathbb{Z})$