100 Cau Hoi Trac Nghiem Chuong I GT 11 Luong GiacUploaded by
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100 CAU HOI TRAC NGHIEM CHUONG I GT 11 LUONG GIAC
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100 Cau Hoi Trac Nghiem Chuong I GT 11 Luong Giac
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100 CAU HOI TRAC NGHIEM CHUONG I GT 11 LUONG GIAC
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\[\begin{array}{l}2{\sin ^2}x + m\sin 2x = 2m\\ \Leftrightarrow 1 - \cos 2x + m\sin 2x = 2m\\ \Leftrightarrow m\sin 2x - \cos 2x = 2m - 1\end{array}\]
Để phương trình vô nghiệm khi: \[{\left[ {2m - 1} \right]^2} > {m^2} + 1\]\[ \Leftrightarrow 3{m^2} - 4m > 0 \Leftrightarrow m \in \left[ { - \infty ;0} \right] \cup \left[ {\frac{4}{3}; + \infty } \right]\]