Giải phương trình sau; Cos2x-√3sin2x=1 Question Giải phương trình sau; Cos2x-√3sin2x=1 in progress 0 Môn Toán Maris 4 years 2020-11-02T07:56:27+00:00 2020-11-02T07:56:27+00:00 2 Answers 249 views 0
Answers ( )
Đáp án:
$\left[\begin{array}{l}x= k\pi\\x = -\dfrac{\pi}{3}+ k\pi\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\cos2x-\sqrt3\sin2x=1$
$\Leftrightarrow \dfrac{1}{2}\cos2x – \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}$
$\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3}$
$\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = \dfrac{\pi}{3} + k2\pi\\2x + \dfrac{\pi}{3} = -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= k\pi\\x = -\dfrac{\pi}{3} +k\pi\end{array}\right.\quad (k\in \Bbb Z)$
$cos2x-\sqrt{3}sin2x=1$
↔ $2(\frac{1}{2}.cos2x-\frac{\sqrt{3}}{2}.sin2x=1$
↔$sin\frac{π}{3}.cos2x-sin2x.cos{π}{6}=\frac{1}{2}$ (
↔$sin(2x-\frac{π}{3})=sin\frac{π}{6}$
↔$\left[ \begin{array}{l}2x-\frac{π}{3}=\frac{π}{6}+k2π\\2x-\frac{π}{3}=π-\frac{π}{6}+k2π\end{array} \right.$
↔$\left[ \begin{array}{l}x=\frac{π}{4}+kπ\\2x=\frac{7π}{12}+kπ\end{array} \right.(k∈Z)$
Chúc bạn học tốt!