1. cos(2x + $\frac{2\pi}{3}$ ) + sin(x + $\frac{\pi}{3}$ ) = 0

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = -\dfrac{\pi}{2} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\cos\left(2x + \dfrac{2\pi}{3}\right) + \sin\left(x + \dfrac{\pi}{3}\right) = 0$

$\Leftrightarrow \cos\left[2.\left(x + \dfrac{\pi}{3}\right)\right] + \sin\left(x + \dfrac{\pi}{3}\right) = 0$

$\Leftrightarrow 2\sin^2\left(x + \dfrac{\pi}{3}\right) – \sin\left(x + \dfrac{\pi}{3}\right) -1 = 0$

$\Leftrightarrow \left[\begin{array}{l}\sin\left(x + \dfrac{\pi}{3}\right) = 1\\\sin\left(x + \dfrac{\pi}{3}\right) = -\dfrac{1}{2}\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{2} + k2\pi\\x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{3} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = -\dfrac{\pi}{2} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$