giải các pt sau: a. sin ²x - cos ² = 0 b. tanx - 3cotx = 1
2 câu trả lời
Giải thích các bước giải:
$a.$ $sin^{2}x-cos^{2}x=0$
$⇔-cos2x=0$
$⇔cos2x=0$
$⇔2x=\frac{\pi}{2}+k\pi$
$⇒x=\frac{\pi}{4}+\frac{k\pi}{2}$ $,k∈Z$
$b.$ $tanx-3cotx=1$ $,(x≠\frac{k\pi}{2}, k∈Z)$
$⇔tanx-\frac{3}{tanx}=1$
$⇔\frac{tan^{2}x-tanx-3}{tanx}=0$
$⇔tan^{2}x-tanx-3=0$
$⇔\left[ \begin{array}{l}tanx=\frac{1+\sqrt[]{13}}{12}\\tanx=\frac{1-\sqrt[]{13}}{12}\end{array} \right.$ $⇒\left[ \begin{array}{l}x=arctan(\frac{1+\sqrt[]{13}}{12})+k\pi\\x=arctan(\frac{1-\sqrt[]{13}}{12})+k\pi\end{array}, k∈Z\right.$
#andy
\[\begin{array}{l}
si{n^2}x - co{s^2}x = 0\\
\Rightarrow \left( {sinx - cosx} \right)\left( {sinx + cosx} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
sinx - cosx = 0\\
sinx + cosx = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sqrt 2 .sin\left( {x - \frac{\pi }{4}} \right){\rm{ = }}0\\
\sqrt 2 .sin\left( {x{\rm{ }} + {\rm{ }}\frac{\pi }{4}} \right){\rm{ }} = {\rm{ }}0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
sin\left( {x - \frac{\pi }{4}} \right) = 0\\
sin\left( {x{\rm{ }} + {\rm{ }}\frac{\pi }{4}} \right){\rm{ }} = {\rm{ }}0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x{\rm{ }} - {\rm{ }}\frac{\pi }{4}{\rm{ }} = {\rm{ }}k\pi \\
\;x{\rm{ }} + {\rm{ }}\frac{\pi }{4}{\rm{ }} = {\rm{ }}k\pi
\end{array} \right.\\
\Rightarrow x{\rm{ }} = {\rm{ }}\frac{\pi }{4}{\rm{ }} + {\rm{ }}\frac{{k\pi }}{2}\,k \in {\rm Z}\\
\end{array}\]
Câu hỏi trong lớp
Xem thêm
