giải pt 10/ $2(cosx+ √3sinx)cosx=cosx- √3sinx+1$ 10/07/2021 Bởi Aaliyah giải pt 10/ $2(cosx+ √3sinx)cosx=cosx- √3sinx+1$
Đáp án: $\left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$ Giải thích các bước giải: $2(\cos x +\sqrt3\sin x)\cos x = \cos x – \sqrt3\sin x + 1$ $\Leftrightarrow 2\cos^2x – 1 + 2\sqrt3\sin x\cos x = \cos x – \sqrt3\sin x$ $\Leftrightarrow \cos2x + \sqrt3\sin2x = \cos x – \sqrt3\sin x$ $\Leftrightarrow \dfrac{1}{2}\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}\cos x -\dfrac{\sqrt3}{2}\sin x$ $\Leftrightarrow \cos\left(2x -\dfrac{\pi}{3}\right) = \cos\left(x +\dfrac{\pi}{3}\right)$ $\Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{3} = x + \dfrac{\pi}{3} + k2\pi\\2x – \dfrac{\pi}{3} =- x -\dfrac{\pi}{3} + k2\pi\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$ Bình luận
$2\cos^2x+2\sqrt3\sin x\cos x -1=\cos x-\sqrt3\sin x$ $\Leftrightarrow \cos2x+\sqrt3\sin2x=\cos x-\sqrt3\sin x$ $\Leftrightarrow 2\sin(2x+\dfrac{\pi}{6})=2\sin(\dfrac{\pi}{6}-x)$ $\Leftrightarrow \sin(2x+\dfrac{\pi}{6})=\sin(\dfrac{\pi}{6}-x)$ $+)TH1: 2x+\dfrac{\pi}{6}=\dfrac{\pi}{6}-x+k2\pi\Leftrightarrow x=\dfrac{k2\pi}{3}$ $+)TH2: 2x+\dfrac{\pi}{6}=x+\dfrac{5\pi}{6}+k2\pi\Leftrightarrow x=\dfrac{2\pi}{3}+k2\pi$ Bình luận
Đáp án:
$\left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$2(\cos x +\sqrt3\sin x)\cos x = \cos x – \sqrt3\sin x + 1$
$\Leftrightarrow 2\cos^2x – 1 + 2\sqrt3\sin x\cos x = \cos x – \sqrt3\sin x$
$\Leftrightarrow \cos2x + \sqrt3\sin2x = \cos x – \sqrt3\sin x$
$\Leftrightarrow \dfrac{1}{2}\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}\cos x -\dfrac{\sqrt3}{2}\sin x$
$\Leftrightarrow \cos\left(2x -\dfrac{\pi}{3}\right) = \cos\left(x +\dfrac{\pi}{3}\right)$
$\Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{3} = x + \dfrac{\pi}{3} + k2\pi\\2x – \dfrac{\pi}{3} =- x -\dfrac{\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x= k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
$2\cos^2x+2\sqrt3\sin x\cos x -1=\cos x-\sqrt3\sin x$
$\Leftrightarrow \cos2x+\sqrt3\sin2x=\cos x-\sqrt3\sin x$
$\Leftrightarrow 2\sin(2x+\dfrac{\pi}{6})=2\sin(\dfrac{\pi}{6}-x)$
$\Leftrightarrow \sin(2x+\dfrac{\pi}{6})=\sin(\dfrac{\pi}{6}-x)$
$+)TH1: 2x+\dfrac{\pi}{6}=\dfrac{\pi}{6}-x+k2\pi\Leftrightarrow x=\dfrac{k2\pi}{3}$
$+)TH2: 2x+\dfrac{\pi}{6}=x+\dfrac{5\pi}{6}+k2\pi\Leftrightarrow x=\dfrac{2\pi}{3}+k2\pi$