Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
2\sin x.\cos x - 1 = 0\\
\Leftrightarrow \sin 2x - 1 = 0\\
\Leftrightarrow \sin 2x = 1\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
16.\sin x.\cos x.\cos 2x.\cos 4x - \sqrt 3 = 0\\
\Leftrightarrow 8.\left( {2\sin x.\cos x} \right).\cos 2x.\cos 4x - \sqrt 3 = 0\\
\Leftrightarrow 8.\sin 2x.\cos 2x.\cos 4x - \sqrt 3 = 0\\
\Leftrightarrow 4.\left( {2\sin 2x.\cos 2x} \right).\cos 4x - \sqrt 3 = 0\\
\Leftrightarrow 4.\sin 4x.\cos 4x - \sqrt 3 = 0\\
\Leftrightarrow 2.\left( {2\sin 4x.\cos 4x} \right) - \sqrt 3 = 0\\
\Leftrightarrow 2.\sin 8x - \sqrt 3 = 0\\
\Leftrightarrow \sin 8x = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
8x = \dfrac{\pi }{3} + k2\pi \\
8x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
