\(sin\left(2x-\dfrac{\Omega}{4}\right)=sin\left(x+\dfrac{3}{4}\Omega\right)\)
=>\(\left[{}\begin{matrix}2x-\dfrac{\Omega}{4}=x+\dfrac{3}{4}\Omega+k2\Omega\\2x-\dfrac{\Omega}{4}=\Omega-x-\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x-x=\dfrac{3}{4}\Omega+\dfrac{\Omega}{4}+k2\Omega=\Omega+k2\Omega\\3x=\Omega-\dfrac{3}{4}\Omega+\dfrac{\Omega}{4}+k2\Omega=\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\Omega+k2\Omega\\x=\dfrac{1}{6}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)
\(x\in\left(0;\Omega\right)\)
=>\(\left[{}\begin{matrix}\Omega+k2\Omega\in\left(0;\Omega\right)\\\dfrac{1}{6}\Omega+\dfrac{k2\Omega}{3}\in\left(0;\Omega\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2k+1\in\left(0;1\right)\\\dfrac{2k}{3}+\dfrac{1}{6}\in\left(0;1\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}k\in\left(-\dfrac{1}{2};0\right)\\\dfrac{2}{3}k\in\left(-\dfrac{1}{6};\dfrac{5}{6}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left(-\dfrac{1}{2};0\right)\\k\in\left(-\dfrac{1}{4};\dfrac{5}{4}\right)\end{matrix}\right.\Leftrightarrow k\in\left\{0;1\right\}\)
Khi k=0 thì \(x=\dfrac{1}{6}\Omega+0\cdot\dfrac{2\Omega}{3}=\dfrac{1}{6}\Omega\)
Khi k=1 thì \(x=\dfrac{1}{6}\Omega+\dfrac{2\Omega}{3}=\dfrac{5}{6}\Omega\)
=>NGhiệm lớn nhất trên khoảng (0;pi) là \(\dfrac{5}{6}\Omega\)
=>Đúng
