Giai các pt sau 1) 2sinxcosx-1=0 2)16 sinxcosxcos2xcos4x- căn3=0 13/07/2021 Bởi Adalynn Giai các pt sau 1) 2sinxcosx-1=0 2)16 sinxcosxcos2xcos4x- căn3=0
Giải thích các bước giải: Ta có: \(\begin{array}{l}1,\\2\sin x.\cos x – 1 = 0\\ \Leftrightarrow \sin 2x – 1 = 0\\ \Leftrightarrow \sin 2x = 1\\ \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\2,\\16.\sin x.\cos x.\cos 2x.\cos 4x – \sqrt 3 = 0\\ \Leftrightarrow 8.\left( {2\sin x.\cos x} \right).\cos 2x.\cos 4x – \sqrt 3 = 0\\ \Leftrightarrow 8.\sin 2x.\cos 2x.\cos 4x – \sqrt 3 = 0\\ \Leftrightarrow 4.\left( {2\sin 2x.\cos 2x} \right).\cos 4x – \sqrt 3 = 0\\ \Leftrightarrow 4.\sin 4x.\cos 4x – \sqrt 3 = 0\\ \Leftrightarrow 2.\left( {2\sin 4x.\cos 4x} \right) – \sqrt 3 = 0\\ \Leftrightarrow 2.\sin 8x – \sqrt 3 = 0\\ \Leftrightarrow \sin 8x = \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}8x = \dfrac{\pi }{3} + k2\pi \\8x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\\x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
2\sin x.\cos x – 1 = 0\\
\Leftrightarrow \sin 2x – 1 = 0\\
\Leftrightarrow \sin 2x = 1\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
16.\sin x.\cos x.\cos 2x.\cos 4x – \sqrt 3 = 0\\
\Leftrightarrow 8.\left( {2\sin x.\cos x} \right).\cos 2x.\cos 4x – \sqrt 3 = 0\\
\Leftrightarrow 8.\sin 2x.\cos 2x.\cos 4x – \sqrt 3 = 0\\
\Leftrightarrow 4.\left( {2\sin 2x.\cos 2x} \right).\cos 4x – \sqrt 3 = 0\\
\Leftrightarrow 4.\sin 4x.\cos 4x – \sqrt 3 = 0\\
\Leftrightarrow 2.\left( {2\sin 4x.\cos 4x} \right) – \sqrt 3 = 0\\
\Leftrightarrow 2.\sin 8x – \sqrt 3 = 0\\
\Leftrightarrow \sin 8x = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
8x = \dfrac{\pi }{3} + k2\pi \\
8x = \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)