giúp mk với ạ :)) thanks các bn nhiều November 4, 2020 by Cherry giúp mk với ạ :)) thanks các bn nhiều
Đáp án: \(\left[ \begin{array}{l}x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{2} + \dfrac{{k2\pi }}{3}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\sin 3x – \sqrt 3 \cos 3x = – 1\\ \to \dfrac{1}{2}\sin 3x – \dfrac{{\sqrt 3 }}{2}\cos 3x = – \dfrac{1}{2}\\ \to \sin 3x.\cos \dfrac{\pi }{3} – \sin \dfrac{\pi }{3}\cos 3x = – \dfrac{1}{2}\\ \to \sin \left( {3x – \dfrac{\pi }{3}} \right) = – \dfrac{1}{2}\\ \to \left[ \begin{array}{l}3x – \dfrac{\pi }{3} = – \dfrac{\pi }{6} + k2\pi \\3x – \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l}3x = \dfrac{\pi }{6} + k2\pi \\3x = \dfrac{{3\pi }}{2} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{2} + \dfrac{{k2\pi }}{3}\end{array} \right.\left( {k \in Z} \right)\end{array}\) Reply
Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{2} + \dfrac{{k2\pi }}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sin 3x – \sqrt 3 \cos 3x = – 1\\
\to \dfrac{1}{2}\sin 3x – \dfrac{{\sqrt 3 }}{2}\cos 3x = – \dfrac{1}{2}\\
\to \sin 3x.\cos \dfrac{\pi }{3} – \sin \dfrac{\pi }{3}\cos 3x = – \dfrac{1}{2}\\
\to \sin \left( {3x – \dfrac{\pi }{3}} \right) = – \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
3x – \dfrac{\pi }{3} = – \dfrac{\pi }{6} + k2\pi \\
3x – \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{3\pi }}{2} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{2} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)