Tổng các nghiệm của phương trình sin(2x-π/3)=-1 trong khoảng (0;5π) bằng 02/08/2021 Bởi Genesis Tổng các nghiệm của phương trình sin(2x-π/3)=-1 trong khoảng (0;5π) bằng
$\sin(2x-\dfrac{\pi}{3})=-1$ $\Leftrightarrow 2x-\dfrac{\pi}{3}=\dfrac{-\pi}{2}+k2\pi$ $\Leftrightarrow x=\dfrac{-\pi}{12}+k\pi$ $0<x<5\pi$ $\Rightarrow 0<\dfrac{-\pi}{12}+k\pi<5\pi$ $\Leftrightarrow 0,083< k<5,08$ $k\in\mathbb{Z}\Rightarrow k\in\{1;2;3;4;5\}$ $\Rightarrow x=\dfrac{11\pi}{12}; \dfrac{23\pi}{12};\dfrac{35\pi}{12}; \dfrac{47\pi}{12}; \dfrac{59\pi}{12}$ $\to$ tổng 5 nghiệm: $\dfrac{175\pi}{12}$ Bình luận
Đáp án: $\sum = \dfrac{175\pi}{12}$ Giải thích các bước giải: $\begin{array}{l}\sin\left(2x – \dfrac{\pi}{3} \right) = -1\\ \Leftrightarrow 2x – \dfrac{\pi}{3} = – \dfrac{\pi}{2} + k2\pi\\ \Leftrightarrow 2x = – \dfrac{\pi}{6} + k2\pi\\ \Leftrightarrow x = – \dfrac{\pi}{12} + k\pi\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ x \in (0;5\pi)\\ \Leftrightarrow 0 < – \dfrac{\pi}{12} + k\pi < 5\pi\\ \Leftrightarrow \dfrac{1}{2} < k < \dfrac{61}{12}\\ \Rightarrow k = \left\{1;2;3;4;5\right\} \Rightarrow x = \left\{\dfrac{11\pi}{12};\dfrac{23\pi}{12};\dfrac{35\pi}{12};\dfrac{47\pi}{12};\dfrac{59\pi}{12}\right\}\\ \sum x = \dfrac{(11 + 23 + 35 + 47 + 59)\pi}{12} = \dfrac{175\pi}{12} \end{array}$ Bình luận
$\sin(2x-\dfrac{\pi}{3})=-1$
$\Leftrightarrow 2x-\dfrac{\pi}{3}=\dfrac{-\pi}{2}+k2\pi$
$\Leftrightarrow x=\dfrac{-\pi}{12}+k\pi$
$0<x<5\pi$
$\Rightarrow 0<\dfrac{-\pi}{12}+k\pi<5\pi$
$\Leftrightarrow 0,083< k<5,08$
$k\in\mathbb{Z}\Rightarrow k\in\{1;2;3;4;5\}$
$\Rightarrow x=\dfrac{11\pi}{12}; \dfrac{23\pi}{12};\dfrac{35\pi}{12}; \dfrac{47\pi}{12}; \dfrac{59\pi}{12}$
$\to$ tổng 5 nghiệm: $\dfrac{175\pi}{12}$
Đáp án:
$\sum = \dfrac{175\pi}{12}$
Giải thích các bước giải:
$\begin{array}{l}\sin\left(2x – \dfrac{\pi}{3} \right) = -1\\ \Leftrightarrow 2x – \dfrac{\pi}{3} = – \dfrac{\pi}{2} + k2\pi\\ \Leftrightarrow 2x = – \dfrac{\pi}{6} + k2\pi\\ \Leftrightarrow x = – \dfrac{\pi}{12} + k\pi\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ x \in (0;5\pi)\\ \Leftrightarrow 0 < – \dfrac{\pi}{12} + k\pi < 5\pi\\ \Leftrightarrow \dfrac{1}{2} < k < \dfrac{61}{12}\\ \Rightarrow k = \left\{1;2;3;4;5\right\} \Rightarrow x = \left\{\dfrac{11\pi}{12};\dfrac{23\pi}{12};\dfrac{35\pi}{12};\dfrac{47\pi}{12};\dfrac{59\pi}{12}\right\}\\ \sum x = \dfrac{(11 + 23 + 35 + 47 + 59)\pi}{12} = \dfrac{175\pi}{12} \end{array}$