Question

giúp mình với

Answer 1

a) \(2sin2x-1=0\Rightarrow sin2x=\dfrac{1}{2}=sin\dfrac{\pi}{6}\)

    \(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+k2\pi\\2x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)     Bạn tự tìm x nhé

Answer 2

2)  \(2sin3x+\sqrt{3}=0\Rightarrow sin3x=-\dfrac{\sqrt{3}}{2}=sin\left(-\dfrac{\pi}{3}\right)\)

    \(\Rightarrow\left[{}\begin{matrix}3x=-\dfrac{\pi}{3}+k2\pi\\3x=\pi-\left(-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)  Bạn tự tìm x nhé.