Toán Lớp 11: Giải pt
1) 2cos2x/ 1-sin2x =0
2) cos2x tanx =0
3) sin3x cotx =0
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Toán Lớp 11: Giải pt
1) 2cos2x/ 1-sin2x =0
2) cos2x tanx =0
3) sin3x cotx =0
Comments ( 1 )
1,\\
x = \dfrac{\pi }{4} + \dfrac{{\left( {2l + 1} \right)\pi }}{2}\,\,\,\,\left( {l \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{{\left( {3k + 1} \right)\pi }}{3}\\
x = \dfrac{{\left( {3k + 2} \right)\pi }}{3}\\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)
1,\\
DKXD:\,\,\,1 – \sin 2x \ne 0 \Leftrightarrow \sin 2x \ne 1 \Leftrightarrow 2x \ne \dfrac{\pi }{2} + k2\pi \Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right)\\
\dfrac{{2\cos 2x}}{{1 – \sin 2x}} = 0\\
\Leftrightarrow 2\cos 2x = 0\\
\Leftrightarrow \cos 2x = 0\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
x \ne \dfrac{\pi }{4} + k\pi \Rightarrow x = \dfrac{\pi }{4} + \dfrac{{\left( {2l + 1} \right)\pi }}{2}\,\,\,\,\left( {l \in Z} \right)\\
2,\\
DKXD:\,\,\,\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)\\
\cos 2x.\tan x = 0\\
\Leftrightarrow \cos 2x.\dfrac{{\sin x}}{{\cos x}} = 0\\
\Leftrightarrow \cos 2x.\sin x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
x = k\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = k\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
3,\\
DKXD:\,\,\,\sin x \ne 0 \Leftrightarrow x \ne k\pi \,\,\,\,\left( {k \in Z} \right)\\
\sin 3x.\cot x = 0\\
\Leftrightarrow \sin 3x.\dfrac{{\cos x}}{{\sin x}} = 0\\
\Leftrightarrow \sin 3x.\cos x = 0\\
\Leftrightarrow \sin 3x.\cos x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 3x = 0\\
\cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = k\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
x \ne k\pi \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{\left( {3k + 1} \right)\pi }}{3}\\
x = \dfrac{{\left( {3k + 2} \right)\pi }}{3}\\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)