Giải phương trình \(\frac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \frac{\pi }{4}} \right)}}{{1 + \tan x}} = \frac{1}{{\sqrt 2 }}\cos x\,\,\,\left( 1 \right)\)
Câu 378100: Giải phương trình \(\frac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \frac{\pi }{4}} \right)}}{{1 + \tan x}} = \frac{1}{{\sqrt 2 }}\cos x\,\,\,\left( 1 \right)\)
A. \(S = \left\{ {\frac{{ - \pi }}{6} + k2\pi ;\frac{{7\pi }}{6} + k2\pi ;\,k \in \mathbb{Z} } \right\}\,\)
B. \(S = \left\{ {\frac{\pi }{6} + k2\pi ;\frac{{5\pi }}{6} + k2\pi ;\,k \in \mathbb{Z} } \right\}\,\)
C. \(S = \left\{ {\frac{{ - \pi }}{3} + k2\pi ;\frac{{4\pi }}{3} + k2\pi ;\,k \in \mathbb{Z} } \right\}\,\)
D. \(S = \left\{ {\frac{\pi }{3} + k2\pi ;\frac{{2\pi }}{3} + k2\pi ;\,k \in \mathbb{Z} } \right\}\,\)
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Đáp án : A(8) bình luận (0) lời giải
Giải chi tiết:
\(\frac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \frac{\pi }{4}} \right)}}{{1 + \tan x}} = \frac{1}{{\sqrt 2 }}\cos x\,\,\,\left( 1 \right)\)
ĐK: \(\left\{ \begin{array}{l}\tan x \ne - 1\\\cos x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{{ - \pi }}{4} + k\pi \\x \ne \frac{\pi }{2} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)
\(\begin{array}{l}\left( 1 \right) \Leftrightarrow \left( {1 + \sin x + \cos 2x} \right).sin\left( {x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}.\cos x.\frac{{\cos x + \sin x}}{{\cos x}}\\ \Leftrightarrow \left( {1 + \sin x + \cos 2x} \right).\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \cos x + \sin x\\ \Leftrightarrow \left( {1 + \sin x + \cos 2x} \right).\left( {\sin x + \cos x} \right) = \cos x + \sin x\\ \Leftrightarrow \left( {\sin x + \cos x} \right).\left( {\sin x + \cos 2x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 0\\\sin x + \cos 2x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 0\\\sin x + 1 - 2{\sin ^2}x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt 2 .\sin \left( {x + \frac{\pi }{4}} \right) = 0\\\sin x = 1\\\sin x = \frac{{ - 1}}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k\pi \,\,\left( {ktm} \right)\\x = \frac{\pi }{2} + k2\pi \,\,\,\left( {ktm} \right)\\x = \frac{{ - \pi }}{6} + k2\pi \,\,\left( {tm} \right)\\x = \frac{{7\pi }}{6} + k2\pi \,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Vậy \(S = \left\{ {\frac{{ - \pi }}{6} + k2\pi ;\frac{{7\pi }}{6} + k2\pi ;\,k \in \mathbb{Z} } \right\}\,\).
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