Toán Lớp 11: Giúp mình với ạ
Đề : giải các phương trình sau
a) cos3x=sinx b) cos (x- 120$^{0}$ )= -sin2x c) cos3x+cos( $\frac{\pi}{4}$+ $\frac{x}{2}$)=0
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Toán Lớp 11: Giúp mình với ạ
Đề : giải các phương trình sau
a) cos3x=sinx b) cos (x- 120$^{0}$ )= -sin2x c) cos3x+cos( $\frac{\pi}{4}$+ $\frac{x}{2}$)=0
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = – 210^\circ + k.360^\circ \\
x = 10^\circ + k.120^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{14}} + \dfrac{{k4\pi }}{7}\\
x = \dfrac{\pi }{2} + \dfrac{{k4\pi }}{5}
\end{array} \right.\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
\cos 3x = \sin x\\
\Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} – x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} – x + k2\pi \\
3x = x – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k2\pi \\
2x = – \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = – \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\cos \left( {x – 120^\circ } \right) = – \sin 2x\\
\Leftrightarrow \cos \left( {x – 120^\circ } \right) = \sin \left( { – 2x} \right)\\
\Leftrightarrow \cos \left( {x – 120^\circ } \right) = \cos \left[ {90^\circ – \left( { – 2x} \right)} \right]\\
\Leftrightarrow \cos \left( {x – 120^\circ } \right) = \cos \left( {90^\circ + 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x – 120^\circ = 90^\circ + 2x + k.360^\circ \\
120^\circ – x = 90^\circ + 2x + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– x = 210^\circ + k.360^\circ \\
– 3x = – 30^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 210^\circ + k.360^\circ \\
x = 10^\circ + k.120^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\cos 3x + \cos \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) = 0\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) = – \cos 3x\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) = \cos \left( {\pi – 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{4} + \dfrac{x}{2} = \pi – 3x + k2\pi \\
\dfrac{\pi }{4} + \dfrac{x}{2} = 3x – \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{7x}}{2} = \dfrac{{3\pi }}{4} + k2\pi \\
– \dfrac{{5x}}{2} = – \dfrac{{5\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{14}} + \dfrac{{k4\pi }}{7}\\
x = \dfrac{\pi }{2} + \dfrac{{k4\pi }}{5}
\end{array} \right.\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)