tìm nghiệm âm lớn nhất của pt sau : a) sin4x + cos5x = 0 b) $sin^{2}$ 2x + cos$^{2}$ 5x =1 05/09/2021 Bởi Cora tìm nghiệm âm lớn nhất của pt sau : a) sin4x + cos5x = 0 b) $sin^{2}$ 2x + cos$^{2}$ 5x =1
Đáp án: $\begin{array}{l}a)\sin 4x + \cos 5x = 0\\ \Rightarrow \sin 4x = – \cos 5x\\ \Rightarrow \sin 4x = cos\left( {5x + \pi } \right)\\ \Rightarrow \sin 4x = \sin \left( {\dfrac{\pi }{2} – 5x – \pi } \right)\\ \Rightarrow \left[ \begin{array}{l}4x = – 5x – \dfrac{\pi }{2} + k2\pi \\4x = \pi + 5x + \dfrac{\pi }{2} + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\\x = \dfrac{{ – 3\pi }}{2} – k2\pi \end{array} \right.\\ \Rightarrow x = – \dfrac{\pi }{{18}}\\b){\sin ^2}2x + {\cos ^2}5x = 1\\ \Rightarrow {\sin ^2}2x + {\cos ^2}5x = {\sin ^2}5x + {\cos ^2}5x\\ \Rightarrow {\sin ^2}2x = {\sin ^2}5x\\ \Rightarrow \left[ \begin{array}{l}\sin 2x = sin5x\\sin2x = – sin5x\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}2x = 5x + k2\pi \\2x = \pi – 5x + k2\pi \\2x = 5x + \pi + k2\pi \\2x = – 5x + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{{k2\pi }}{3}\\x = \dfrac{\pi }{7} + \dfrac{{k2\pi }}{7}\\x = \dfrac{{ – \pi }}{3} – \dfrac{{k2\pi }}{3}\\x = \dfrac{{k2\pi }}{7}\end{array} \right.\\ \Rightarrow x = – \dfrac{\pi }{7}\left( {khi:k = – 1} \right)\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\sin 4x + \cos 5x = 0\\
\Rightarrow \sin 4x = – \cos 5x\\
\Rightarrow \sin 4x = cos\left( {5x + \pi } \right)\\
\Rightarrow \sin 4x = \sin \left( {\dfrac{\pi }{2} – 5x – \pi } \right)\\
\Rightarrow \left[ \begin{array}{l}
4x = – 5x – \dfrac{\pi }{2} + k2\pi \\
4x = \pi + 5x + \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\\
x = \dfrac{{ – 3\pi }}{2} – k2\pi
\end{array} \right.\\
\Rightarrow x = – \dfrac{\pi }{{18}}\\
b){\sin ^2}2x + {\cos ^2}5x = 1\\
\Rightarrow {\sin ^2}2x + {\cos ^2}5x = {\sin ^2}5x + {\cos ^2}5x\\
\Rightarrow {\sin ^2}2x = {\sin ^2}5x\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = sin5x\\
sin2x = – sin5x
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = 5x + k2\pi \\
2x = \pi – 5x + k2\pi \\
2x = 5x + \pi + k2\pi \\
2x = – 5x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{7} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{ – \pi }}{3} – \dfrac{{k2\pi }}{3}\\
x = \dfrac{{k2\pi }}{7}
\end{array} \right.\\
\Rightarrow x = – \dfrac{\pi }{7}\left( {khi:k = – 1} \right)
\end{array}$
Đáp án:
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