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Đáp án:
\(\left[ \begin{array}{l} x = m2\pi \\ x = \frac{\pi }{2} + l2\pi \end{array} \right.\,\,\,\,voi\,\,m,\,\,\,l,\,\,k \in Z\,\,\,thoa\,\,man\,\,\left[ \begin{array}{l} m \ne \frac{k}{4}\\ l \ne \frac{{k - 1}}{4} \end{array} \right..\)
Giải thích các bước giải: \(\begin{array}{l} \frac{1}{{\sin x}} + \frac{1}{{\cos x}} = \frac{2}{{\sin 2x}}\,\,\,\left( * \right)\\ DK:\,\,\sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}.\\ \left( * \right) \Leftrightarrow \frac{1}{{\sin x}} + \frac{1}{{\cos x}} = \frac{2}{{2\sin x\cos x}}\\ \Leftrightarrow \frac{1}{{\sin x}} + \frac{1}{{\cos x}} = \frac{1}{{\sin x\cos x}}\\ \Leftrightarrow \cos x + \sin x = 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 1\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \frac{\pi }{4} + m2\pi \\ x + \frac{\pi }{4} = \frac{{3\pi }}{4} + l2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = m2\pi \\ x = \frac{\pi }{2} + l2\pi \end{array} \right.\,\,\,\left( {m,\,\,l \in Z} \right)\\ \Rightarrow Pt\,\,co\,\,nghiem \Leftrightarrow \left[ \begin{array}{l} m2\pi \ne \frac{{k\pi }}{2}\\ \frac{\pi }{2} + l2\pi \ne \frac{{k\pi }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m \ne \frac{k}{4}\\ l2\pi \ne \frac{{\left( {k - 1} \right)\pi }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m \ne \frac{k}{4}\\ l \ne \frac{{k - 1}}{4} \end{array} \right..\\ Vay\,\,\,pt\,\,co\,\,nghiem:\,\,\left[ \begin{array}{l} x = m2\pi \\ x = \frac{\pi }{2} + l2\pi \end{array} \right.\,\,\,\,voi\,\,m,\,\,\,l,\,\,k \in Z\,\,\,thoa\,\,man\,\,\left[ \begin{array}{l} m \ne \frac{k}{4}\\ l \ne \frac{{k - 1}}{4} \end{array} \right.. \end{array}\)
