Đáp án:
\(S = \left\{\dfrac{\pi}{24} + k\dfrac{\pi}{2};\ \dfrac{5\pi}{24} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \cos^2\left(2x+\dfrac{\pi}{4}\right)= \dfrac14\\
\Leftrightarrow \dfrac{1 + \cos\left(4x + \dfrac{\pi}{2}\right)}{2} = \dfrac14\\
\Leftrightarrow \cos\left(4x + \dfrac{\pi}{2}\right) = -\dfrac12\\
\Leftrightarrow \sin4x = \dfrac12\\
\Leftrightarrow \left[\begin{array}{l}4x = \dfrac{\pi}{6} + k2\pi\\4x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{24} + k\dfrac{\pi}{2}\\x = \dfrac{5\pi}{24} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{24} + k\dfrac{\pi}{2};\ \dfrac{5\pi}{24} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
