Đáp án:
a) $\left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$
b) $\left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
a) $\sin(2x + 15^o) = \dfrac{\sqrt2}{2}$
$\Leftrightarrow \sin(2x + 15^o) = \sin45^o$
$\Leftrightarrow \left[\begin{array}{l}2x +15^o = 45^o + k.360^o\\2x + 15^o = 135^o + k.360^o\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= 15^o + k.180^o\\x= 60^o + k.180^o\end{array}\right.\quad (k \in \Bbb Z)$
b) $\sin\left(2x - \dfrac{\pi}{5}\right) = -\dfrac{\sqrt3}{2}$
$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{5}\right) = \sin\left(-\dfrac{\pi}{3}\right)$
$\Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{5} = -\dfrac{\pi}{3} + k2\pi\\2x - \dfrac{\pi}{5}= \dfrac{4\pi}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= -\dfrac{\pi}{15}+ k\pi\\x= \dfrac{23\pi}{30}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$
