Đáp án:
a) $x =\dfrac{\pi}{9} + k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $\cos5x\cos2x +\sqrt3\sin3x = 2 -\sin5x\sin2x$
$\Leftrightarrow \cos5x\cos2x +\sin5x\sin2x + \sqrt3\sin3x = 2$
$\Leftrightarrow \cos3x +\sqrt3\sin3x = 2$
$\Leftrightarrow \dfrac12\cos3x +\dfrac{\sqrt3}{2}\sin3x = 1$
$\Leftrightarrow \cos\left(3x -\dfrac{\pi}{3}\right) = 1$
$\Leftrightarrow 3x -\dfrac{\pi}{3} = k2\pi$
$\Leftrightarrow x =\dfrac{\pi}{9} + k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)$
b) $\cos^3x -\sin^3x + 1 = 0$
$\Leftrightarrow \cos^3x - \sin^3x + \cos^2x +\sin^2x = 0$
$\Leftrightarrow \cos^2x(\cos x +1) +\sin^2x(1-\sin x)=0$
$\Leftrightarrow (1-\sin^2x)(\cos x +1) +(1-\cos^2x)(1-\sin x) = 0$
$\Leftrightarrow (1-\sin x)(1+\sin x)(\cos x +1) +(1-\cos x)(1+\cos x)(1-\sin x) = 0$
$\Leftrightarrow (1-\sin x)(1+\cos x)(1 +\sin x + 1 -\cos x) = 0$
$\Leftrightarrow \Leftrightarrow (1-\sin x)(1+\cos x)(2 +\sin x -\cos x) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 1\\\cos x = -1\\\sin x -\cos x = -2\quad \text{(vô nghiệm)}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
