a) \(2sim\left(2x-\dfrac{\pi}{3}\right)-1=0\Rightarrow sin\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}=sin\dfrac{\pi}{6}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.
d) \(\sqrt{3}sinx-cosx=\sqrt{2}\) \(\Rightarrow2sin\left(x-\dfrac{\pi}{6}\right)=\sqrt{2}\)
\(\Rightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}=sin\dfrac{\pi}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.
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