Question

cho \(x\in\left(\frac{-\pi}{2};\frac{\pi}{2}\right)\) và t/m \(cos2x+3sinx-2=0\) . tính \(M=sin2x.cosx\)

Answer 1

\(cos2x+3sinx-2=0\)

\(\Leftrightarrow1-2sin^2x+3sinx-2=0\)

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

Do \(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow\left\{{}\begin{matrix}-1< sinx< 1\\0< cosx\le1\end{matrix}\right.\)

\(\Rightarrow sinx=\frac{1}{2}\) \(\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{\sqrt{3}}{2}\)

\(\Rightarrow M=sin2x.cosx=2sinx.cos^2x=2.\frac{1}{2}.\left(\frac{\sqrt{3}}{2}\right)^2=...\)