Toán căn 3.(sin2x – cosx) + cos2x – sinx=0 giúp mk với 03/12/2021 By Mackenzie căn 3.(sin2x – cosx) + cos2x – sinx=0 giúp mk với
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: $\sqrt3(\sin2x -\cos x) +\cos2x -\sin x = 0$ $\to \sqrt3\sin2x +\cos2x = \sqrt3\cos x +\sin x$ $\to \dfrac{\sqrt3}{2}\sin2x +\dfrac12\cos2x = \dfrac{\sqrt3}{2}\cos x +\dfrac12\sin x$ $\to \sin\left(2x +\dfrac{\pi}{6}\right) =\sin\left(x +\dfrac{\pi}{3}\right)$ $\to \left[\begin{array}{l}2x +\dfrac{\pi}{6}=x +\dfrac{\pi}{3} + k2\pi\\2x +\dfrac{\pi}{6}=\dfrac{2\pi}{3} -x+ k2\pi\end{array}\right.$ $\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{6}+ k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$ Trả lời
`~rai~` $\begin{array}{I}\sqrt{3}(sin2x-cosx)+cos2x-sinx=0\\\Leftrightarrow \sqrt{3}sin2x+cos2x=sinx+\sqrt{3}cosx\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\\\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(x+\dfrac{\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3} (k\in \mathbb{Z})\end{array}\right.\end{array}$ Trả lời
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:
$\sqrt3(\sin2x -\cos x) +\cos2x -\sin x = 0$
$\to \sqrt3\sin2x +\cos2x = \sqrt3\cos x +\sin x$
$\to \dfrac{\sqrt3}{2}\sin2x +\dfrac12\cos2x = \dfrac{\sqrt3}{2}\cos x +\dfrac12\sin x$
$\to \sin\left(2x +\dfrac{\pi}{6}\right) =\sin\left(x +\dfrac{\pi}{3}\right)$
$\to \left[\begin{array}{l}2x +\dfrac{\pi}{6}=x +\dfrac{\pi}{3} + k2\pi\\2x +\dfrac{\pi}{6}=\dfrac{2\pi}{3} -x+ k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{6}+ k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
`~rai~`
$\begin{array}{I}\sqrt{3}(sin2x-cosx)+cos2x-sinx=0\\\Leftrightarrow \sqrt{3}sin2x+cos2x=sinx+\sqrt{3}cosx\\\Leftrightarrow \dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\\\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(x+\dfrac{\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3} (k\in \mathbb{Z})\end{array}\right.\end{array}$