điều kiện : \(x\ne\dfrac{\pi}{2}+k2\pi\)
pt \(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\\sqrt{3}sinx+cosx=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{4}+k\pi\\sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{4}+k\pi\\sin\left(x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{6}\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{4}+k\pi\\x=k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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