Đáp án:
$2$ nghiệm
Giải thích các bước giải:
$\begin{array}{l}2\sin\left(x + \dfrac{\pi}{3}\right) = 1\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{3}\right) = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ 0 \leq x \leq 2\pi\\ \Leftrightarrow \left[\begin{array}{l}0 \leq -\dfrac{\pi}{6} + k2\pi \leq 2\pi\\0 \leq \dfrac{\pi}{2} + k2\pi \leq 2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{1}{12} \leq k \leq \dfrac{13}{12}\\-\dfrac{1}{4} \leq k \leq \dfrac{3}{4}\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}k = 1\\k = 0\end{array}\right.\quad (k \in \Bbb Z)\\ \Rightarrow \left[\begin{array}{l}x = \dfrac{11\pi}{6} \\x = \dfrac{\pi}{2} \end{array}\right.\\ \text{Vậy có 2 nghiệm thỏa mãn đề bài} \end{array}$
