cos5x+2sinxcosx+2sin3xsin2x=0
⇔cos5x+2sinxcosx+\(\dfrac{1}{2}\)(cosx-cos5x)*2=0
⇔cos5x+2sinxcosx+cosx-cos5x=0
⇔cosx(1+2sinx)=0
⇔cosx=0 hoặc sinx=\(\dfrac{-1}{2}\)
⇔x=\(\dfrac{\Pi}{2}+k\Pi\) hoặc x=\(\dfrac{-1}{6}\Pi+k2\Pi\) hoặc x=\(\dfrac{7}{6}\Pi+k2\Pi\) với k∈Z
Lời giải:
\(\cos 5x+2\sin x\cos x+2\sin 3x\sin 2x=0\)
\(\Leftrightarrow \cos (3x+2x)+2\sin x\cos x+2\sin 3x\sin 2x=0\)
\(\Leftrightarrow \cos 3x\cos 2x-\sin 3x\sin 2x+2\sin x\cos x+2\sin3x\sin 2x=0\)
\(\Leftrightarrow (\cos 3x\cos 2x+\sin 3x\sin 2x)+2\sin x\cos x=0\)
\(\Leftrightarrow \cos (3x-2x)+2\sin x\cos x=0\)
\(\Leftrightarrow \cos x(1+2\sin x)=0\)
\(\Rightarrow \left[\begin{matrix} \cos x=0\\ 1+2\sin x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} \cos x=0\\ \sin x=\frac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=k\pi+\frac{\pi}{2}\\ x=\frac{-\pi}{6}+2k\pi\\ x=\frac{7\pi}{6}+2k\pi\end{matrix}\right.\) (k nguyên)
