Đặt \(t=sinx\) , \(-1\le t\le1\)
Phương trình đã cho trở thành:
\(4t^2-2\left(\sqrt{3}+1\right)t+\sqrt{3}=0\)
\(\Leftrightarrow\left(2t-1\right)\left(2t-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1}{2}\\t=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\) (nhận)
+ Với \(sinx=\dfrac{1}{2}\Rightarrow sinx=sin\dfrac{\pi}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
+ Với \(sinx=\dfrac{\sqrt{3}}{2}\Rightarrow sinx=sin\dfrac{\pi}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
Vậy ....
