Toán Lớp 11: Giải phương trình sau
a, sinx + $\sqrt[]{3}$cosx = -1
b, 3sin5x + 4cos5x = 5
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Toán Lớp 11: Giải phương trình sau
a, sinx + $\sqrt[]{3}$cosx = -1
b, 3sin5x + 4cos5x = 5
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = \dfrac{\pi }{{10}} – \dfrac{{\arccos \dfrac{3}{5}}}{5} + \dfrac{{k2\pi }}{5}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
\sin x + \sqrt 3 \cos x = – 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x = – \dfrac{1}{2}\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6} = – \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right) = \cos \dfrac{{2\pi }}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{6} = \dfrac{{2\pi }}{3} + k2\pi \\
x – \dfrac{\pi }{6} = – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
3\sin 5x + 4\cos 5x = 5\\
\Leftrightarrow \dfrac{3}{5}\sin 5x + \dfrac{4}{5}\cos 5x = 1\\
\Leftrightarrow \sin 5x.\cos \alpha + \cos 5x.\sin \alpha = 1\\
\left( {\cos \alpha = \dfrac{3}{5};\,\,\sin \alpha = \dfrac{4}{5};\,\,{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1} \right)\\
\Leftrightarrow \sin \left( {5x + \alpha } \right) = 1\\
\Leftrightarrow 5x + \alpha = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 5x = \dfrac{\pi }{2} – \alpha + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{{10}} – \dfrac{\alpha }{5} + \dfrac{{k2\pi }}{5}\\
\Leftrightarrow x = \dfrac{\pi }{{10}} – \dfrac{{\arccos \dfrac{3}{5}}}{5} + \dfrac{{k2\pi }}{5}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)