Toán Tổng tất cả các nghiệm của phương trình √3 sinx – cosx = 0 trên đoạn [- π; π] 24/11/2021 By Brielle Tổng tất cả các nghiệm của phương trình √3 sinx – cosx = 0 trên đoạn [- π; π]
Đáp án: $-\dfrac{2\pi}{3}$ Giải thích các bước giải: $\begin{array}{l}\quad \sqrt3\sin x-\cos x = 0\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin x – \dfrac12\cos x =0\\ \Leftrightarrow \sin\left(x – \dfrac{\pi}{6}\right) =0\\ \Leftrightarrow x – \dfrac{\pi}{6} = k\pi\\ \Leftrightarrow x = \dfrac{\pi}{6} + k\pi\quad (k\in\Bbb Z)\\ \text{Ta có:}\\ \quad -\pi \leq x \leq \pi\\ \to -\pi \leq \dfrac{\pi}{6} + k\pi \leq \pi\\ \to -\dfrac76 \leq k \leq \dfrac56\\ Do\,\,k\in\Bbb Z\\ nên\,\,k \in \{-1;0\}\\ \to \left[\begin{array}{l}x =- \dfrac{5\pi}{6}\\x = \dfrac{\pi}{6} \end{array}\right.\\ \to S = -\dfrac{5\pi}{6} + \dfrac{\pi}{6} = -\dfrac{2\pi}{3} \end{array}$ Trả lời
`~rai~` $\begin{array}{I}\sqrt{3}sinx-cosx=0\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}x=0\\\Leftrightarrow cos\dfrac{\pi}{6}sinx-sin\dfrac{\pi}{6}cosx=0\\\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=0\\\Leftrightarrow x-\dfrac{\pi}{6}=k\pi\\\Leftrightarrow x=\dfrac{\pi}{6}+k\pi(k\in\mathbb{Z})\\\text{Ta có:}-\pi\le x\le \pi\\\Leftrightarrow -\pi\le \dfrac{\pi}{6}+k\pi\le \pi\\\Leftrightarrow -\dfrac{\pi}{6}\le k\pi\le \dfrac{5\pi}{6}\\\Leftrightarrow -\dfrac{7}{6}\le k\le \dfrac{5}{6}\\\text{Vì k}\in\mathbb{Z}\Rightarrow k\in\{-1;0\}.\\\Leftrightarrow x_1=-\dfrac{5\pi}{6};x_2=\dfrac{\pi}{6}\\\text{Tổng các nghiệm là:}-\dfrac{5\pi}{6}+\dfrac{\pi}{6}=-\dfrac{2\pi}{3}. \end{array}$ Trả lời
Đáp án:
$-\dfrac{2\pi}{3}$
Giải thích các bước giải:
$\begin{array}{l}\quad \sqrt3\sin x-\cos x = 0\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin x – \dfrac12\cos x =0\\ \Leftrightarrow \sin\left(x – \dfrac{\pi}{6}\right) =0\\ \Leftrightarrow x – \dfrac{\pi}{6} = k\pi\\ \Leftrightarrow x = \dfrac{\pi}{6} + k\pi\quad (k\in\Bbb Z)\\ \text{Ta có:}\\ \quad -\pi \leq x \leq \pi\\ \to -\pi \leq \dfrac{\pi}{6} + k\pi \leq \pi\\ \to -\dfrac76 \leq k \leq \dfrac56\\ Do\,\,k\in\Bbb Z\\ nên\,\,k \in \{-1;0\}\\ \to \left[\begin{array}{l}x =- \dfrac{5\pi}{6}\\x = \dfrac{\pi}{6} \end{array}\right.\\ \to S = -\dfrac{5\pi}{6} + \dfrac{\pi}{6} = -\dfrac{2\pi}{3} \end{array}$
`~rai~`
$\begin{array}{I}\sqrt{3}sinx-cosx=0\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}x=0\\\Leftrightarrow cos\dfrac{\pi}{6}sinx-sin\dfrac{\pi}{6}cosx=0\\\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=0\\\Leftrightarrow x-\dfrac{\pi}{6}=k\pi\\\Leftrightarrow x=\dfrac{\pi}{6}+k\pi(k\in\mathbb{Z})\\\text{Ta có:}-\pi\le x\le \pi\\\Leftrightarrow -\pi\le \dfrac{\pi}{6}+k\pi\le \pi\\\Leftrightarrow -\dfrac{\pi}{6}\le k\pi\le \dfrac{5\pi}{6}\\\Leftrightarrow -\dfrac{7}{6}\le k\le \dfrac{5}{6}\\\text{Vì k}\in\mathbb{Z}\Rightarrow k\in\{-1;0\}.\\\Leftrightarrow x_1=-\dfrac{5\pi}{6};x_2=\dfrac{\pi}{6}\\\text{Tổng các nghiệm là:}-\dfrac{5\pi}{6}+\dfrac{\pi}{6}=-\dfrac{2\pi}{3}. \end{array}$