Đáp án:
\(S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad 1+\cot2x = \dfrac{1 - \cos2x}{\sin^22x}\qquad (*)\\
ĐK:\sin2x \ne 0 \Leftrightarrow x \ne \dfrac{n\pi}{2}\\
(*) \Leftrightarrow \sin^22x + \sin2x.\cos2x = 1 - \cos2x\\
\Leftrightarrow (\sin^22x - 1) + \sin2x.\cos2x + \cos2x = 0\\
\Leftrightarrow (\sin2x - 1)(\sin2x + 1) + \cos2x(\sin2x + 1) = 0\\
\Leftrightarrow (\sin2x + 1)(\sin2x + \cos2x - 1) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin2x = -1\\\sin2x + \cos2x = 1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\sin2x = -1\\\sin\left(2x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}2x = -\dfrac{\pi}{2} + k2\pi\\2x + \dfrac{\pi}{4} =\dfrac{\pi}{4} + k2\pi\\2x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x= k\pi\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x = k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Kết hợp ĐKXĐ ta được:}\ x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
