Đáp án:

$2$ nghiệm thỏa yêu cầu bài toán

Giải thích các bước giải:

$\begin{array}{l}\sqrt2\cos\left(x + \dfrac{\pi}{3}\right) = 1\\ \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{3} = -\dfrac{\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{-\pi}{12} + k2\pi\\x = -\dfrac{7\pi}{12} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ 0 \leq x \leq 2\pi\\ \Leftrightarrow \left[\begin{array}{l}0 \leq \dfrac{-\pi}{12} + k2\pi \leq 2\pi\\0 \leq -\dfrac{7\pi}{12} + k2\pi \leq 2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{1}{24} \leq k \leq \dfrac{25}{24}\\\dfrac{7}{24}\leq k \leq \dfrac{31}{24}\end{array}\right.\quad (k \in \Bbb Z)\\ \Leftrightarrow \left[\begin{array}{l}k = 1\\k = 1\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x = \dfrac{23\pi}{12}\\x = \dfrac{17\pi}{12}\end{array}\right.\\ \end{array}$

Phương trình có 2 nghiệm.