Question

Tìm nghiệm của pt: \(2sin^2x-3sinx+1=0\) sao cho \(0\le x< \pi\)

Answer 1

\(2sin^2x-3sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Do 0\(\le\) x<\(\pi\) nên :

+ \(0\le\frac{\pi}{2}+k2\pi< \pi\) <=> \(-\frac{1}{4}\le k< \frac{1}{4}\) => \(x_1=\frac{\pi}{2}\)

+ \(0\le\frac{\pi}{6}+k2\pi< \pi\Leftrightarrow-\frac{1}{12}\le k< \frac{5}{12}\) => \(x_2=\frac{\pi}{6}\)

+ \(0\le\frac{5\pi}{6}+k2\pi< \pi\Leftrightarrow-\frac{5}{12}\le k< \frac{1}{12}\) => x₃ = \(\frac{5\pi}{6}\)