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:

$2(\cos x +\sqrt3\sin x)\cos x = \cos x - \sqrt3\sin x + 1$

$\Leftrightarrow 2\cos^2x - 1 + 2\sqrt3\sin x\cos x = \cos x - \sqrt3\sin x$

$\Leftrightarrow \cos2x + \sqrt3\sin2x = \cos x - \sqrt3\sin x$

$\Leftrightarrow \dfrac{1}{2}\cos2x + \dfrac{\sqrt3}{2}\sin2x = \dfrac{1}{2}\cos x -\dfrac{\sqrt3}{2}\sin x$

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

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

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