Trường hợp này nên khai triển hết ra:
\(\Leftrightarrow\sin3x.\cos\frac{5\pi}{6}-\cos3x.\sin\frac{5\pi}{6}+\cos3x.\cos\frac{3\pi}{4}-\sin3x.\sin\frac{3\pi}{4}=0\)
\(\Leftrightarrow\frac{-\sqrt{3}}{2}\sin3x-\frac{1}{2}\cos3x-\frac{\sqrt{2}}{2}\cos3x-\frac{\sqrt{2}}{2}\sin3x=0\)
\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)\sin3x=-\left(1+\sqrt{2}\right)\cos3x\)
\(\Rightarrow\sin3x=-\frac{1+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\cos3x\)
\(\Rightarrow\left\{{}\begin{matrix}\sin3x=-\frac{1+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\cos3x\\\sin^23x+\cos^23x=1\end{matrix}\right.\)
Hệ phương trình 2 ẩn, tự giải nha :)
\(\Leftrightarrow sin\left(3x-\frac{5\pi}{6}\right)=-cos\left(3x+\frac{3\pi}{4}\right)=sin\left(3x+\frac{3\pi}{4}-\frac{\pi}{2}\right)\)
\(\Leftrightarrow sin\left(3x-\frac{5\pi}{6}\right)=sin\left(3x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow3x-\frac{5\pi}{6}=\pi-3x-\frac{\pi}{4}+k2\pi\)
\(\Leftrightarrow6x=\frac{19\pi}{12}+k2\pi\)
\(\Leftrightarrow x=\frac{19\pi}{72}+\frac{k\pi}{3}\)
