Toán Lớp 12: giải các pt sau:
a. sin ²x – cos ² = 0
b. tanx – 3cotx = 1
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Toán Lớp 12: giải các pt sau:
a. sin ²x – cos ² = 0
b. tanx – 3cotx = 1
Comments ( 2 )
si{n^2}x – co{s^2}x = 0\\
\Rightarrow \left( {sinx – cosx} \right)\left( {sinx + cosx} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
sinx – cosx = 0\\
sinx + cosx = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sqrt 2 .sin\left( {x – \frac{\pi }{4}} \right){\rm{ = }}0\\
\sqrt 2 .sin\left( {x{\rm{ }} + {\rm{ }}\frac{\pi }{4}} \right){\rm{ }} = {\rm{ }}0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
sin\left( {x – \frac{\pi }{4}} \right) = 0\\
sin\left( {x{\rm{ }} + {\rm{ }}\frac{\pi }{4}} \right){\rm{ }} = {\rm{ }}0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x{\rm{ }} – {\rm{ }}\frac{\pi }{4}{\rm{ }} = {\rm{ }}k\pi \\
\;x{\rm{ }} + {\rm{ }}\frac{\pi }{4}{\rm{ }} = {\rm{ }}k\pi
\end{array} \right.\\
\Rightarrow x{\rm{ }} = {\rm{ }}\frac{\pi }{4}{\rm{ }} + {\rm{ }}\frac{{k\pi }}{2}\,k \in {\rm Z}\\
\end{array}\]
$⇔-cos2x=0$
$⇔\frac{tan^{2}x-tanx-3}{tanx}=0$