Đáp án:
$x = k\pi\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\cos\left(x + \dfrac{\pi}{3}\right) + \cos\left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{\cos x}\qquad (*)$
$ĐKXĐ:\, x \ne \dfrac{\pi}{2} + n\pi$
$(*)\Leftrightarrow 2\cos\left(\dfrac{x + \dfrac{\pi}{3} + x - \dfrac{\pi}{3}}{2}\right).\cos\left(\dfrac{x + \dfrac{\pi}{3} - x +\dfrac{\pi}{3}}{2}\right)=\dfrac{1}{\cos x}$
$\Leftrightarrow \cos x\cos\dfrac{\pi}{3} = \dfrac{1}{2\cos x}$
$\Leftrightarrow \cos^2x.\dfrac{1}{2} = \dfrac{1}{2}$
$\Leftrightarrow \cos^2x = 1$
$\Leftrightarrow \cos x = \pm 1$
$\Leftrightarrow x = k\pi\quad (k\in\Bbb Z)$
