Giải các phương trình sau
1. $2\sin 4x+1=0$.
2. $3\cos x-2=0$.
3. $3\tan \left(x+\frac{\pi }{6}\right)-\sqrt{3}=0$.
4. $\cot (3x+1)+1=0$.
5. $(\sin 3x+1)(2\cos x-\sqrt{3})=0$.
6. $\sin 7x+\cos 2x=0$.

(AI giải đề cho tôi)
Bài 3. Giải các phương trình lượng giác:

1) $2\sin 4x+1=0\iff \sin 4x=-\frac{1}{2}$
$\iff 4x=-\frac{\pi }{6}+2k\pi$ hoặc $4x=\pi +\frac{\pi }{6}+2k\pi =\frac{7\pi }{6}+2k\pi$
$\iff x=-\frac{\pi }{24}+\frac{k\pi }{2}$ hoặc $x=\frac{7\pi }{24}+\frac{k\pi }{2}$, $k\in \mathbb{Z}$.

2) $3\cos x-2=0\iff \cos x=\frac{2}{3}$ (thỏa mãn vì $\left|\frac{2}{3}\right|\le 1$)
$\iff x=\pm \arccos \frac{2}{3}+2k\pi$, $k\in \mathbb{Z}$.

3) $3\tan \left(x+\frac{\pi }{6}\right)-\sqrt{3}=0\iff \tan \left(x+\frac{\pi }{6}\right)=\frac{\sqrt{3}}{3}=\tan \frac{\pi }{6}$
$\iff x+\frac{\pi }{6}=\frac{\pi }{6}+k\pi \iff x=k\pi$, $k\in \mathbb{Z}$.

4) $\cot (3x+1)+1=0\iff \cot (3x+1)=-1=\cot \left(-\frac{\pi }{4}\right)$ (với $3x+1\ne k\pi$)
$\iff 3x+1=-\frac{\pi }{4}+k\pi \iff x=-\frac{1}{3}-\frac{\pi }{12}+\frac{k\pi }{3}$, $k\in \mathbb{Z}$.

5) $(\sin 3x+1)(2\cos x-\sqrt{3})=0$
• $\sin 3x=-1\iff 3x=-\frac{\pi }{2}+2k\pi \iff x=-\frac{\pi }{6}+\frac{2k\pi }{3}$.
• $\cos x=\frac{\sqrt{3}}{2}\iff x=\pm \frac{\pi }{6}+2k\pi$.
Vậy $x=-\frac{\pi }{6}+\frac{2k\pi }{3}$ hoặc $x=\pm \frac{\pi }{6}+2k\pi$, $k\in \mathbb{Z}$.

6) $\sin 7x+\cos 2x=0\iff \sin 7x=-\cos 2x=\sin \left(2x-\frac{\pi }{2}\right)$
• Trường hợp 1: $7x=2x-\frac{\pi }{2}+2k\pi \iff 5x=-\frac{\pi }{2}+2k\pi \iff x=-\frac{\pi }{10}+\frac{2k\pi }{5}$.
• Trường hợp 2: $7x=\pi -\left(2x-\frac{\pi }{2}\right)+2k\pi =\frac{3\pi }{2}-2x+2k\pi \iff 9x=\frac{3\pi }{2}+2k\pi \iff x=\frac{\pi }{6}+\frac{2k\pi }{9}$.
Vậy $x=-\frac{\pi }{10}+\frac{2k\pi }{5}$ hoặc $x=\frac{\pi }{6}+\frac{2k\pi }{9}$, $k\in \mathbb{Z}$.

Kiểm tra nhanh: (1) $\sin \left(4\cdot \left(-\frac{\pi }{24}\right)\right)=\sin \left(-\frac{\pi }{6}\right)=-\frac{1}{2}$ đúng; (3) với $x=k\pi$, $\tan \left(k\pi +\frac{\pi }{6}\right)=\frac{\sqrt{3}}{3}$ đúng; (6) với $x=\frac{\pi }{6}$: $\sin \frac{7\pi }{6}+\cos \frac{\pi }{3}=-\frac{1}{2}+\frac{1}{2}=0$ đúng.