Câu 26 trang 32 SGK Đại số và Giải tích 11 Nâng cao

LG a

\(\cos 3x = \sin 2x\)

Lời giải chi tiết:

\(\eqalign{& \cos 3x = \sin 2x \cr& \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} – 2x} \right)\cr&\Leftrightarrow \cos 3x – \cos \left( {{\pi \over 2} – 2x} \right) = 0 \cr & \Leftrightarrow  – 2\sin \left( {\frac{{3x + \frac{\pi }{2} – 2x}}{2}} \right)\sin \left( {\frac{{3x – \frac{\pi }{2} + 2x}}{2}} \right) = 0\cr&\Leftrightarrow – 2\sin \left( {{x \over 2} + {\pi \over 4}} \right)\sin \left( {{{5x} \over 2} – {\pi \over 4}} \right) = 0 \cr} \)

\( \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = 0\\
\sin \left( {\frac{{5x}}{2} – \frac{\pi }{4}} \right) = 0
\end{array} \right.\)

\( \Leftrightarrow \left[ {\matrix{{{x \over 2} + {\pi \over 4} = k\pi } \\ {{{5x} \over 2} – {\pi \over 4} = k\pi } \cr} } \right.\\ \Leftrightarrow \left[ {\matrix{{x = – {\pi \over 2} + k2\pi } \\ {x = {\pi \over {10}} + k{{2\pi } \over 5}} } } \right. ,k\in Z\)

LG b

\(\sin (x – 120˚) – \cos 2x = 0\)

Lời giải chi tiết:

\(\eqalign{& \sin \left( {x – 120^\circ } \right) – \cos 2x = 0 \cr& \Leftrightarrow \cos \left( {{{90}^0} – x + {{120}^0}} \right) – \cos 2x = 0\cr&\Leftrightarrow \cos \left( {210^\circ – x} \right) – \cos 2x = 0 \cr &  \Leftrightarrow  – 2\sin \left( {\frac{{{{210}^0} – x + 2x}}{2}} \right)\sin \left( {\frac{{{{210}^0} – x – 2x}}{2}} \right) = 0\cr&\Leftrightarrow – 2\sin \left( {{x \over 2} + 105^\circ } \right)\sin \left( {105^\circ – {{3x} \over 2}} \right) = 0 \cr} \)

\( \Leftrightarrow \left[ \begin{array}{l}
\sin \left( {\frac{x}{2} + {{105}^0}} \right) = 0\\
\sin \left( {{{105}^0} – \frac{{3x}}{2}} \right) = 0
\end{array} \right.\)

\( \Leftrightarrow \left[ {\matrix{{{x \over 2} + 105^\circ = k180^\circ } \\ {105^\circ – {{3x} \over 2} = k180^\circ } \cr} } \right. \\\Leftrightarrow \left[ {\matrix{{x = – 210^\circ + k360^\circ } \\ {x = 70^\circ – k120^\circ } \cr} } \right. ,k\in Z\)

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