$\cos 2x-\sqrt{3}\sin x+2=0\Leftrightarrow -2\sin^{2}x-\sqrt{3}\sin x+3=0\Leftrightarrow \left\{ \begin{array}{l} \sin x=\frac{\sqrt{3}}{2}\\ \sin x=-\sqrt{3}(loại) \end{array} \right.$$\cos 3x+\cos x+2\sin ^{2}x-2=0\Leftrightarrow 4\cos ^{3}x-2\cos x+2(\sin ^{2}x-1)=0\Leftrightarrow \cos x(4\cos^{2}x+2\cos x-2)=0\Leftrightarrow \left\{ \begin{array}{l} \cos x=0,\cos x=-1\\ \cos x=\frac{1}{2} \end{array} \right.$$\sin x+9\sin \frac{x}{3}-8=0\Leftrightarrow 12\sin \frac{x}{3}-4\sin ^{3}\frac{x}{3}-8=0\Leftrightarrow \left\{ \begin{array}{l} \sin \frac{x}{3}=1\\ \sin \frac{x}{3}=-2(loại) \end{array} \right.$$2\cos x^{2}\frac{6x}{5}+2\cos \frac{8x}{5}-\cos \frac{4x}{5}-3=0$$\Leftrightarrow \cos \frac{12x}{5}+\cos \frac{4x}{5}+2\cos \frac{8x}{5}-2(\cos\frac{4x}{5}+1)=0$$\Leftrightarrow 2\cos \frac{8x}{5}(\cos \frac{4x}{5}+1)-2(\cos \frac{4x}{5}+1)=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos \frac{8x}{5}=1\\ \cos \frac{4x}{5}=-1 \end{array} \right.$5.$2\cos 3x-\sin (x-\frac{\pi }{6})-1=0\Leftrightarrow -2\cos (3x+\pi )+\cos ((x-\frac{\pi }{6})+\frac{\pi }{2})-1=0$Đặt $t=x+\frac{\pi }{3}$ thì PT $\Leftrightarrow -2\cos 3t+\cos t-1=0\Leftrightarrow -8\cos ^{3}t+7\cos t-1=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos t=-1\\ \cos t=\frac{2+\sqrt{6}}{4},\cos t=\frac{2-\sqrt{6}}{4} \end{array} \right.$

$\cos 2x-\sqrt{3}\sin x+2=0\Leftrightarrow -2\sin^{2}x-\sqrt{3}\sin x+3=0\Leftrightarrow \left\{ \begin{array}{l} \sin x=\frac{\sqrt{3}}{2}\\ \sin x=-\sqrt{3}(loại) \end{array} \right.$$\cos 3x+\cos x+2\sin ^{2}x-2=0\Leftrightarrow 4\cos ^{3}x-2\cos x+2(\sin ^{2}x-1)=0\Leftrightarrow \cos x(4\cos^{2}x+2\cos x-2)=0\Leftrightarrow \left\{ \begin{array}{l} \cos x=0,\cos x=-1\\ \cos x=\frac{1}{2} \end{array} \right.$$\sin x+9\sin \frac{x}{3}-8=0\Leftrightarrow 12\sin \frac{x}{3}-4\sin ^{3}\frac{x}{3}-8=0\Leftrightarrow \left\{ \begin{array}{l} \sin \frac{x}{3}=1\\ \sin \frac{x}{3}=-2(loại) \end{array} \right.$$2\cos x^{2}\frac{6x}{5}+2\cos \frac{8x}{5}-\cos \frac{4x}{5}-3=0$$\Leftrightarrow \cos \frac{12x}{5}+\cos \frac{4x}{5}+2\cos \frac{8x}{5}-2(\cos\frac{4x}{5}+1)=0$$\Leftrightarrow 2\cos \frac{8x}{5}(\cos \frac{4x}{5}+1)-2(\cos \frac{4x}{5}+1)=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos \frac{8x}{5}=1\\ \cos \frac{4x}{5}=-1 \end{array} \right.$5.$2\cos 3x-\sin (x-\frac{\pi }{6})-1=0\Leftrightarrow -2\cos (3x+\pi )+\cos ((x-\frac{\pi }{6})+\frac{\pi }{2})-1=0$Đặt $t=x+\frac{\pi }{3}$ thì PT $\Leftrightarrow -2\cos 3t+\cos t-1=0\Leftrightarrow -8\cos ^{3}t+7\cos t-1=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos t=-1\\ \cos t=\frac{2+\sqrt{6}}{4}(loại),\cos t=\frac{2-\sqrt{6}}{4} \end{array} \right.$

$\cos 2x-\sqrt{3}\sin x+2=0\Leftrightarrow -2\sin^{2}x-\sqrt{3}\sin x+3=0\Leftrightarrow \left\{ \begin{array}{l} \sin x=\frac{\sqrt{3}}{2}\\ \sin x=-\sqrt{3}(loại) \end{array} \right.$$\cos 3x+\cos x+2\sin ^{2}x-2=0\Leftrightarrow 4\cos ^{3}x-2\cos x+2(\sin ^{2}x-1)=0\Leftrightarrow \cos x(4\cos^{2}x+2\cos x-2)=0\Leftrightarrow \left\{ \begin{array}{l} \cos x=0,\cos x=-1\\ \cos x=\frac{1}{2} \end{array} \right.$$\sin x+9\sin \frac{x}{3}-8=0\Leftrightarrow 12\sin \frac{x}{3}-4\sin ^{3}\frac{x}{3}-8=0\Leftrightarrow \left\{ \begin{array}{l} \sin \frac{x}{3}=1\\ \sin \frac{x}{3}=-2(loại) \end{array} \right.$$2\cos x^{2}\frac{6x}{5}+2\cos \frac{8x}{5}-\cos \frac{4x}{5}-3=0$$\Leftrightarrow \cos \frac{12x}{5}+\cos \frac{4x}{5}+2\cos \frac{8x}{5}-2(\cos\frac{4x}{5}+1)=0$$\Leftrightarrow 2\cos \frac{8x}{5}(\cos \frac{4x}{5}+1)-2(\cos \frac{4x}{5}+1)=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos \frac{8x}{5}=1\\ \cos \frac{4x}{5}=-1 \end{array} \right.$câu cuối giải chưa ra nhưng mong cậu vẫn cho tớ 2000 sò nha,chắc 4/5 là được rồi nhỉ.

$\cos 2x-\sqrt{3}\sin x+2=0\Leftrightarrow -2\sin^{2}x-\sqrt{3}\sin x+3=0\Leftrightarrow \left\{ \begin{array}{l} \sin x=\frac{\sqrt{3}}{2}\\ \sin x=-\sqrt{3}(loại) \end{array} \right.$$\cos 3x+\cos x+2\sin ^{2}x-2=0\Leftrightarrow 4\cos ^{3}x-2\cos x+2(\sin ^{2}x-1)=0\Leftrightarrow \cos x(4\cos^{2}x+2\cos x-2)=0\Leftrightarrow \left\{ \begin{array}{l} \cos x=0,\cos x=-1\\ \cos x=\frac{1}{2} \end{array} \right.$$\sin x+9\sin \frac{x}{3}-8=0\Leftrightarrow 12\sin \frac{x}{3}-4\sin ^{3}\frac{x}{3}-8=0\Leftrightarrow \left\{ \begin{array}{l} \sin \frac{x}{3}=1\\ \sin \frac{x}{3}=-2(loại) \end{array} \right.$$2\cos x^{2}\frac{6x}{5}+2\cos \frac{8x}{5}-\cos \frac{4x}{5}-3=0$$\Leftrightarrow \cos \frac{12x}{5}+\cos \frac{4x}{5}+2\cos \frac{8x}{5}-2(\cos\frac{4x}{5}+1)=0$$\Leftrightarrow 2\cos \frac{8x}{5}(\cos \frac{4x}{5}+1)-2(\cos \frac{4x}{5}+1)=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos \frac{8x}{5}=1\\ \cos \frac{4x}{5}=-1 \end{array} \right.$5.$2\cos 3x-\sin (x-\frac{\pi }{6})-1=0\Leftrightarrow -2\cos (3x+\pi )+\cos ((x-\frac{\pi }{6})+\frac{\pi }{2})-1=0$Đặt$t=x+\frac{\pi }{3}$ thì PT $\Leftrightarrow -2\cos 3t+\cos t-1=0\Leftrightarrow -8\cos ^{3}t+7\cos t-1=0$$\Leftrightarrow \left\{ \begin{array}{l} \cos t=-1\\ \cos t=\frac{2+\sqrt{6}}{4},\cos t=\frac{2-\sqrt{6}}{4} \end{array} \right.$