Question

[Nguyễn thị Phụng]

Phương trình : \(3cot^2x+2\sqrt{2}sin^2x=\left(2+3\sqrt{2}\right)cosx\) có các nghiệm dạng \(x=\alpha+k2\Pi;x=\beta+k2\Pi\) , \(0< \alpha,\beta< \frac{\Pi}{2}\) thì \(\alpha.\beta\) bằng :

A. \(\frac{\Pi^2}{12}\)

B. \(-\frac{\Pi^2}{12}\)

C. \(\frac{7\Pi}{12}\)

D. \(\frac{\Pi^2}{12^2}\)

Trình bày bài giải chi tiết rồi ms chọn đáp án nha các bạn .

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: ...

\(\Leftrightarrow\frac{3cos^2x}{sin^2x}-2cosx+2\sqrt{2}sin^2x-3\sqrt{2}cosx=0\)

\(\Leftrightarrow cosx\left(\frac{3cosx-2sin^2x}{sin^2x}\right)-\sqrt{2}\left(3cosx-2sin^2x\right)=0\)

\(\Leftrightarrow\left(3cosx-2sin^2x\right)\left(\frac{cosx}{sin^2x}-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3cosx-2sin^2x=0\\cosx-\sqrt{2}sin^2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2cos^2x+3cosx-2=0\\\sqrt{2}cos^2x+cosx-\sqrt{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\alpha.\beta=\frac{\pi^2}{12}\)