Question

Giải pt: sin\(\left(2x-\frac{\pi}{6}\right)\)=sin\(\left(\frac{\pi}{6}-x\right)+2\)

Answer 1

Do \(VT=sin\left(2x-\frac{\pi}{6}\right)\le1\)

\(sin\left(\frac{\pi}{6}-x\right)\ge-1\Rightarrow VT=sin\left(\frac{\pi}{6}-x\right)+2\ge-1+2=1\)

\(\Rightarrow VP\ge VT\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)=1\\sin\left(\frac{\pi}{6}-x\right)=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\\\frac{\pi}{6}-x=-\frac{\pi}{2}+l2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=\frac{2\pi}{3}+l2\pi\end{matrix}\right.\) \(\Rightarrow x=\varnothing\)