Toán Lớp 11: Câu 1:cos(3x-60°)= √3/2
Câu2:giải pt √ 3sinx -cosx = 1
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Toán Lớp 11: Câu 1:cos(3x-60°)= √3/2
Câu2:giải pt √ 3sinx -cosx = 1
Giúp mình mới ạ
Comments ( 1 )
1,\\
\left[ \begin{array}{l}
x = 30^\circ + k.120^\circ \\
x = 10^\circ + k.120^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
1,\\
\cos \left( {3x – 60^\circ } \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \cos \left( {3x – 60^\circ } \right) = \cos 30^\circ \\
\Leftrightarrow \left[ \begin{array}{l}
3x – 60^\circ = 30^\circ + k.360^\circ \\
3x – 60^\circ = – 30^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 90^\circ + k.360^\circ \\
3x = 30^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 30^\circ + k.120^\circ \\
x = 10^\circ + k.120^\circ
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sqrt 3 \sin x – \cos x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x – \dfrac{1}{2}\cos x = \dfrac{1}{2}\\
\Leftrightarrow \sin x.\cos \dfrac{\pi }{6} – \cos x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x – \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)