2.Phương trình đã cho tương đương với: $\sin9x-\sqrt3\cos9x=\sin7x-\sqrt3\cos7x$$\Leftrightarrow \dfrac{1}{2}\sin9x-\dfrac{\sqrt3}{2}\cos9x=\dfrac{1}{2}\sin7x-\dfrac{\sqrt3}{2}\cos7x$$\Leftrightarrow \sin(9x-\dfrac{\pi}{3})=\sin(7x-\dfrac{\pi}{3})$$\Leftrightarrow \left[\begin{array}{l}9x-\dfrac{\pi}{3}=7x-\dfrac{\pi}{3}+k2\pi\\9x-\dfrac{\pi}{3}=\dfrac{4\pi}{3}-7x+k2\pi\end{array}\right.,k\in\mathbb{Z}$$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{5\pi}{48}+k\dfrac{\pi}{8}\end{array}\right.,k\in\mathbb{Z}$
2.Phương trình đã cho tương đương với: $\sin9x-\sqrt3\cos9x=\sin7x-\sqrt3\cos7x$$\Leftrightarrow \dfrac{1}{2}\sin9x-\dfrac{\sqrt3}{2}\cos9x=\dfrac{1}{2}\sin7x-\dfrac{\sqrt3}{2}\cos7x$$\Leftrightarrow \sin(9x-\dfrac{\pi}{6})=\sin(7x-\dfrac{\pi}{6})$$\Leftrightarrow \left[\begin{array}{l}9x-\dfrac{\pi}{6}=7x-\dfrac{\pi}{6}+k2\pi\\9x-\dfrac{\pi}{6}=\dfrac{7\pi}{6}-7x+k2\pi\end{array}\right.,k\in\mathbb{Z}$$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{8}\end{array}\right.,k\in\mathbb{Z}$
2.Phương trình đã cho tương đương với: $\sin9x-\sqrt3\cos9x=\sin7x-\sqrt3\cos7x$$\Leftrightarrow \dfrac{1}{2}\sin9x-\dfrac{\sqrt3}{2}\cos9x=\dfrac{1}{2}\sin7x-\dfrac{\sqrt3}{2}\cos7x$$\Leftrightarrow \sin(9x-\dfrac{\pi}{
3})=\sin(7x-\dfrac{\pi}{
3})$$\Leftrightarrow \left[\begin{array}{l}9x-\dfrac{\pi}{
3}=7x-\dfrac{\pi}{
3}+k2\pi\\9x-\dfrac{\pi}{
3}=\dfrac{
4\pi}{
3}-7x+k2\pi\end{array}\right.,k\in\mathbb{Z}$$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{
5\pi}{
48}+k\dfrac{\pi}{8}\end{array}\right.,k\in\mathbb{Z}$