Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt 3 \sin x - \sin x.\cos x = 0\\
\Leftrightarrow \sin x.\left( {\sqrt 3 - \cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = \sqrt 3 \,\,\,\,\left( {L,\,\,\, - 1 \le \cos x \le 1} \right)
\end{array} \right.\\
\Leftrightarrow \sin x = 0\\
\Leftrightarrow x = k\pi \\
b,\\
{\sin ^2}x - \sqrt 3 \sin x.\cos x + 2{\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x - \sqrt 3 \sin x.\cos x + 2{\cos ^2}x = {\sin ^2}x + {\cos ^2}x\\
\Leftrightarrow - \sqrt 3 \sin x.\cos x + {\cos ^2}x = 0\\
\Leftrightarrow \cos x.\left( { - \sqrt 3 \sin x + \cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
- \sqrt 3 \sin x + \cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {x - \dfrac{\pi }{6}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x - \dfrac{\pi }{6} = k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k\pi
\end{array} \right.\\
c,\\
2{\sin ^3}x - {\sin ^2}x.\cos x + 2\sin x.{\cos ^2}x - {\cos ^3}x = 0\\
\Leftrightarrow {\sin ^2}x.\left( {2\sin x - \cos x} \right) + {\cos ^2}x.\left( {2\sin x - \cos x} \right) = 0\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right).\left( {2\sin x - \cos x} \right) = 0\\
\Leftrightarrow 1.\left( {2\sin x - \cos x} \right) = 0\\
\Leftrightarrow 2\sin x - \cos x = 0\\
\Leftrightarrow 2\sin x = \cos x\\
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{1}{2}\\
\Leftrightarrow \tan x = \dfrac{1}{2}\\
\Leftrightarrow x = \arctan \dfrac{1}{2} + k\pi
\end{array}\)
