Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài
Ẩn danh
Nguyễn Lê Phước Thịnh
16 tháng 7 2024 lúc 9:34

10: \(\sqrt{3}\cdot tan\left(2x+\dfrac{\Omega}{3}\right)=-3\)

=>\(tan\left(2x+\dfrac{\Omega}{3}\right)=-\sqrt{3}\)

=>\(2x+\dfrac{\Omega}{3}=-\dfrac{\Omega}{3}+k\Omega\)

=>\(2x=-\dfrac{2}{3}\Omega+k\Omega\)

=>\(x=-\dfrac{1}{3}\Omega+\dfrac{k\Omega}{2}\)

11: \(2\sqrt{2}\cdot sin\left(\dfrac{2x+\Omega}{3}\right)=2\)

=>\(sin\left(\dfrac{2x+\Omega}{3}\right)=\dfrac{1}{\sqrt{2}}\)

=>\(\left[{}\begin{matrix}\dfrac{2x+\Omega}{3}=\dfrac{\Omega}{4}+k2\Omega\\\dfrac{2x+\Omega}{3}=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x+\Omega=\dfrac{3}{4}\Omega+k6\Omega\\2x+\Omega=\dfrac{9}{4}\Omega+k6\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{4}\Omega+k6\Omega\\2x=\dfrac{5}{4}\Omega+k6\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{8}\Omega+k3\Omega\\x=\dfrac{5}{8}\Omega+k3\Omega\end{matrix}\right.\)

12: \(2\sqrt{3}\cdot cos\left(3x+\dfrac{\Omega}{3}\right)-3=0\)

=>\(2\sqrt{3}\cdot cos\left(3x+\dfrac{\Omega}{3}\right)=3\)

=>\(cos\left(3x+\dfrac{\Omega}{3}\right)=\dfrac{3}{2\sqrt{3}}=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}3x+\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\\3x+\dfrac{\Omega}{3}=-\dfrac{\Omega}{6}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\Omega}{6}+k2\Omega-\dfrac{\Omega}{3}=-\dfrac{1}{6}\Omega+k2\Omega\\3x=-\dfrac{\Omega}{6}-\dfrac{\Omega}{3}+k2\Omega=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{18}\Omega+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{6}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)

13: \(6\cdot cos\left(4x+\dfrac{\Omega}{5}\right)+3\sqrt{3}=0\)

=>\(6\cdot cos\left(4x+\dfrac{\Omega}{5}\right)=-3\sqrt{3}\)

=>\(cos\left(4x+\dfrac{\Omega}{5}\right)=-\dfrac{3\sqrt{3}}{6}=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}4x+\dfrac{\Omega}{5}=\dfrac{5}{6}\Omega+k2\Omega\\4x+\dfrac{\Omega}{5}=-\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{19}{30}\Omega+k2\Omega\\4x=-\dfrac{31}{30}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{19}{120}\Omega+\dfrac{k\Omega}{2}\\x=-\dfrac{31}{120}\Omega+\dfrac{k\Omega}{2}\end{matrix}\right.\)

14: \(cos\left(x-\dfrac{\Omega}{3}\right)=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k2\Omega\\x-\dfrac{\Omega}{3}=-\dfrac{\Omega}{3}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\Omega+k2\Omega\\x=k2\Omega\end{matrix}\right.\)


Các câu hỏi tương tự
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết
Pham Trong Bach
Xem chi tiết