Question

4sin²x-2(\(\sqrt{3}\)+1)sĩn+\(\sqrt{3}\)=0

Answer 1

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 ....