Phương trình \({\sin ^2}x + \sin x\cos 4x + {\cos ^2}4x = {3 \over 4}\) có nghiệm là:
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Lời giải:
Báo saiTa có:
\(\eqalign{
& {\sin ^2}x + \sin x\cos 4x + {{{{\cos }^2}4x} \over 4} + {3 \over 4}{\cos ^2}4x = {3 \over 4} \cr
& \Leftrightarrow {\left( {\sin x + {1 \over 2}\cos 4x} \right)^2} = {3 \over 4}\left( {1 - {{\cos }^2}4x} \right)\cr& \Leftrightarrow {\left( {\sin x + {1 \over 2}\cos 4x} \right)^2} = {3 \over 4}{\sin ^2}4x \cr
& \Leftrightarrow \left[ \matrix{
\sin x + {1 \over 2}\cos 4x = {{\sqrt 3 } \over 2}\sin 4x \hfill \cr
\sin x + {1 \over 2}\cos 4x = - {{\sqrt 3 } \over 2}\sin 4x \hfill \cr} \right. \cr& \Leftrightarrow \left[ \matrix{
\cos {\pi \over 6}\sin 4x - \sin {\pi \over 6}\cos 4x = \sin x \hfill \cr
\sin {\pi \over 6}\cos 4x + \cos {\pi \over 6}\sin 4x = \sin \left( { - x} \right) \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
\sin \left( {4x - {\pi \over 6}} \right) = \sin x \hfill \cr
\sin \left( {4x + {\pi \over 6}} \right) = \sin \left( { - x} \right) \hfill \cr} \right. \cr} \)
\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
4x - \frac{\pi }{6} = x + k2\pi \\
4x - \frac{\pi }{6} = \pi - x + k2\pi \\
4x + \frac{\pi }{6} = - x + k2\pi \\
4x + \frac{\pi }{6} = \pi + x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\
x = \frac{{7\pi }}{{30}} + \frac{{k2\pi }}{5}\\
x = - \frac{\pi }{{30}} + \frac{{k2\pi }}{5}\\
x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\)
Vậy \(x = {\pi \over {18}} + k{{2\pi } \over 3},x = {{7\pi } \over {30}} + k{{2\pi } \over 5},\) \(x = - {\pi \over {30}} + k{{2\pi } \over 5},x = {{5\pi } \over {18}} + k{{2\pi } \over 3}\).
