Question

Tìm số nghiệm thuộc \(\left[\frac{-3\pi}{2};-\pi\right]\) của pt

\(\sqrt{3}sinx=cos\left(\frac{3\pi}{2}-2x\right)\)

Answer 1

\(\sqrt{3}sinx=cos\left(\frac{3\pi}{2}-2x\right)\)

\(\Leftrightarrow\sqrt{3}sinx=-cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow\sqrt{3}sinx=-sin2x\)

\(\Leftrightarrow2sinx.cosx+\sqrt{3}sinx=0\)

\(\Leftrightarrow sinx\left(2cosx+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Do\(x\in\left[\frac{-3\pi}{2};-\pi\right]\)

\(\Leftrightarrow x=-\pi;x=\frac{-7\pi}{6};x=\frac{-5\pi}{6}\)