Toán Lớp 11: Mọi người giúp mình giải bài này với
12sinx + 5cosx + 13cos3x=0
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Toán Lớp 11: Mọi người giúp mình giải bài này với
12sinx + 5cosx + 13cos3x=0
Comments ( 1 )
x = \dfrac{\pi }{4} + \dfrac{1}{4}\arccos \dfrac{5}{{13}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{2} – \dfrac{1}{2}\arccos \dfrac{5}{{13}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
12\sin x + 5\cos x + 13\cos 3x = 0\\
\Leftrightarrow 5\cos x + 12\sin x = – 13\cos 3x\\
\Leftrightarrow \dfrac{5}{{13}}\cos x + \dfrac{{12}}{{13}}\sin x = – \cos 3x\\
\Leftrightarrow \cos x.\cos \alpha + \sin x.\sin \alpha = \cos \left( {\pi – 3x} \right)\\
\left( {\cos \alpha = \dfrac{5}{{13}};\,\,\sin \alpha = \dfrac{{12}}{{13}};\,\,\,{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1} \right)\\
\Leftrightarrow \cos \left( {x – \alpha } \right) = \cos \left( {\pi – 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x – \alpha = \pi – 3x + k2\pi \\
x – \alpha = 3x – \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = \pi + \alpha + k2\pi \\
– 2x = – \pi + \alpha + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{1}{4}\alpha + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{2} – \dfrac{1}{2}\alpha + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{1}{4}\arccos \dfrac{5}{{13}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{2} – \dfrac{1}{2}\arccos \dfrac{5}{{13}} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)