Toán Giải pt lượng giác sau sin2x – 2cosx +√3 = √3 sinx 09/07/2021 By Gianna Giải pt lượng giác sau sin2x – 2cosx +√3 = √3 sinx
`sin(2x) – 2cos(x) + \sqrt{3} = \sqrt{3}sin(x)` `⇔ sin(2x) – 2cos(x) + \sqrt{3} – \sqrt{3} sin(x) = 0` `⇔ (1-sin(x))(\sqrt{3}-2cos(x))=0` `⇔`\(\left[ \begin{array}{l}1-sin(x)=0\\\sqrt{3}-2cos(x)=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+2\pi n\\x=\dfrac{\pi}{6}+2\pi n\\x=\dfrac{11\pi}{6}+2\pi n\end{array} \right.\) Trả lời
Đáp án: $\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array}\right. \quad (k \in \Bbb Z)$ Giải thích các bước giải: $\begin{array}{l}\sin2x – 2\cos x + \sqrt3 = \sqrt3\sin x\\ \Leftrightarrow 2\sin x\cos x – 2\cos x + \sqrt3 – \sqrt3\sin x =0\\ \Leftrightarrow 2\cos x(\sin x – 1) – \sqrt3(\sin x – 1) = 0\\ \Leftrightarrow (\sin x – 1)(2\cos x – \sqrt3) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 1\\\cos x = \dfrac{\sqrt3}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array}\right. \quad (k \in \Bbb Z) \end{array}$ Trả lời
`sin(2x) – 2cos(x) + \sqrt{3} = \sqrt{3}sin(x)`
`⇔ sin(2x) – 2cos(x) + \sqrt{3} – \sqrt{3} sin(x) = 0`
`⇔ (1-sin(x))(\sqrt{3}-2cos(x))=0`
`⇔`\(\left[ \begin{array}{l}1-sin(x)=0\\\sqrt{3}-2cos(x)=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+2\pi n\\x=\dfrac{\pi}{6}+2\pi n\\x=\dfrac{11\pi}{6}+2\pi n\end{array} \right.\)
Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array}\right. \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin2x – 2\cos x + \sqrt3 = \sqrt3\sin x\\ \Leftrightarrow 2\sin x\cos x – 2\cos x + \sqrt3 – \sqrt3\sin x =0\\ \Leftrightarrow 2\cos x(\sin x – 1) – \sqrt3(\sin x – 1) = 0\\ \Leftrightarrow (\sin x – 1)(2\cos x – \sqrt3) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 1\\\cos x = \dfrac{\sqrt3}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \pm \dfrac{\pi}{6} + k2\pi\end{array}\right. \quad (k \in \Bbb Z) \end{array}$