Đáp án:

\(\begin{array}{l}
1)\,\,\,\,x = \dfrac{\pi }{3} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
2)\,\,\,\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\,\,\,\,x = k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1)\\
2{\sin ^2}x + \sqrt 3 \sin 2x = 3\\
 \Leftrightarrow 2{\sin ^2}x + \sqrt 3 .2\sin x.\cos x - 3 = 0\\
 \Leftrightarrow 2{\sin ^2}x + 2\sqrt 3 \sin x.\cos x - 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0\\
 \Leftrightarrow 2{\sin ^2}x + 2\sqrt 3 \sin x.\cos x - 3{\sin ^2}x - 3{\cos ^2}x = 0\\
 \Leftrightarrow  - {\sin ^2}x + 2\sqrt 3 \sin x.\cos x - 3{\cos ^2}x = 0\\
 \Leftrightarrow {\sin ^2}x - 2\sqrt 3 \sin x.\cos x + 3{\cos ^2}x = 0\\
 \Leftrightarrow {\sin ^2}x - 2.\sin x.\sqrt 3 \cos x + {\left( {\sqrt 3 \cos x} \right)^2} = 0\\
 \Leftrightarrow {\left( {\sin x - \sqrt 3 \cos x} \right)^2} = 0\\
 \Leftrightarrow \sin x - \sqrt 3 \cos x = 0\\
 \Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{{\sqrt 3 }}{2}\cos x = 0\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{3} - \cos x.\sin \dfrac{\pi }{3} = 0\\
 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{3}} \right) = 0\\
 \Leftrightarrow x - \dfrac{\pi }{3} = k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{3} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right)\\
 \Leftrightarrow \sin 8x - \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\\
 \Leftrightarrow \dfrac{1}{2}\sin 8x - \dfrac{{\sqrt 3 }}{2}\cos 8x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\
 \Leftrightarrow \sin 8x.\cos \dfrac{\pi }{3} - \cos 8x.\sin \dfrac{\pi }{3} = \sin 6x.\cos \dfrac{\pi }{6} + \cos 6x.\sin \dfrac{\pi }{6}\\
 \Leftrightarrow \sin \left( {8x - \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x - \dfrac{\pi }{3} = \pi  - \left( {6x + \dfrac{\pi }{6}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x - \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} - 6x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
14x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\cos x - \sqrt 3 \sin x = 2\cos \left( {\dfrac{\pi }{3} - x} \right)\\
 \Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = \cos \left( {\dfrac{\pi }{3} - x} \right)\\
 \Leftrightarrow \cos x.\cos \dfrac{\pi }{3} - \sin x.\sin \dfrac{\pi }{3} = \cos \left( {\dfrac{\pi }{3} - x} \right)\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{\pi }{3} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{3} - x + k2\pi \\
x + \dfrac{\pi }{3} = x - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = k2\pi \\
0x =  - \dfrac{{2\pi }}{3} + k2\pi \,\,\,\,\left( L \right)
\end{array} \right.\\
 \Rightarrow x = k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)