Question

\(1-5sin\left(\frac{x}{2}\right)+2cos^2\left(\frac{x}{2}\right)=0\)

Answer 1

\(\Leftrightarrow1-5sin\left(\frac{x}{2}\right)+2-2sin^2\left(\frac{x}{2}\right)=0\)

Đặt \(sin\left(\frac{x}{2}\right)=t\Rightarrow\left|t\right|\le1\)

Pt trở thành:

\(-2t^2-5t+3=0\Rightarrow\left[{}\begin{matrix}t=-3\left(l\right)\\t=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(\frac{x}{2}\right)=\frac{1}{2}=sin\left(\frac{\pi}{6}\right)\)

\(\Rightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{6}+k2\pi\\\frac{x}{2}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k4\pi\\x=\frac{5\pi}{3}+k4\pi\end{matrix}\right.\)