Giải phương trình sau; Cos2x-√3sin2x=1

Giải phương trình sau; Cos2x-√3sin2x=1

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  1. 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:

    $\cos2x-\sqrt3\sin2x=1$

    $\Leftrightarrow \dfrac{1}{2}\cos2x – \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}$

    $\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3}$

    $\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = \dfrac{\pi}{3} + k2\pi\\2x + \dfrac{\pi}{3} = -\dfrac{\pi}{3} + k2\pi\end{array}\right.$

    $\Leftrightarrow \left[\begin{array}{l}x= k\pi\\x = -\dfrac{\pi}{3} +k\pi\end{array}\right.\quad (k\in \Bbb Z)$

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  2. $cos2x-\sqrt{3}sin2x=1$

    ↔ $2(\frac{1}{2}.cos2x-\frac{\sqrt{3}}{2}.sin2x=1$

    ↔$sin\frac{π}{3}.cos2x-sin2x.cos{π}{6}=\frac{1}{2}$ (

    ↔$sin(2x-\frac{π}{3})=sin\frac{π}{6}$

    ↔$\left[ \begin{array}{l}2x-\frac{π}{3}=\frac{π}{6}+k2π\\2x-\frac{π}{3}=π-\frac{π}{6}+k2π\end{array} \right.$

    ↔$\left[ \begin{array}{l}x=\frac{π}{4}+kπ\\2x=\frac{7π}{12}+kπ\end{array} \right.(k∈Z)$

    Chúc bạn học tốt!

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