căn 3.(sin2x – cosx) + cos2x – sinx=0 giúp mk với

Question

căn 3.(sin2x – cosx) + cos2x – sinx=0
giúp mk với

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Mackenzie 1 tuần 2021-12-03T23:50:23+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-12-03T23:52:04+00:00

    Đáp án:

    $\left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{6}+ k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$

    Giải thích các bước giải:

    $\sqrt3(\sin2x -\cos x) +\cos2x -\sin x = 0$

    $\to \sqrt3\sin2x +\cos2x = \sqrt3\cos x +\sin x$

    $\to \dfrac{\sqrt3}{2}\sin2x +\dfrac12\cos2x = \dfrac{\sqrt3}{2}\cos x +\dfrac12\sin x$

    $\to \sin\left(2x +\dfrac{\pi}{6}\right) =\sin\left(x +\dfrac{\pi}{3}\right)$

    $\to \left[\begin{array}{l}2x +\dfrac{\pi}{6}=x +\dfrac{\pi}{3} + k2\pi\\2x +\dfrac{\pi}{6}=\dfrac{2\pi}{3} -x+ k2\pi\end{array}\right.$

    $\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{6}+ k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$

    0
    2021-12-03T23:52:06+00:00

    `~rai~`

    $\begin{array}{I}\sqrt{3}(sin2x-cosx)+cos2x-sinx=0\\\Leftrightarrow \sqrt{3}sin2x+cos2x=sinx+\sqrt{3}cosx\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\\\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(x+\dfrac{\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3} (k\in \mathbb{Z})\end{array}\right.\end{array}$

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