\(\Leftrightarrow sin5x+sinx-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
\(sin5x+sinx+2sin^2x=1\)
\(\Leftrightarrow\left(sin5x+sinx\right)-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\end{matrix}\right.\)
Vậy...
\(sin5x+sinx+2sin^2x=1\)
\(\Leftrightarrow2sin3x.cos2x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
