1, sin7x - sinx = cos4x
⇔ 2cos4x . sin3x = cos4x
⇔ \(\left[{}\begin{matrix}cos4x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\)
2, \(\sqrt{3}cos2x\left(2sinx-1\right)+2cosx\left(2sin^2x+1\right)=3sin2x\)
⇔ \(\sqrt{3}cos2x\left(2sinx-1\right)+2cosx\left(2sin^2x-3sinx+1\right)=0\)
⇔ \(\sqrt{3}cos2x\left(2sinx-1\right)+2cosx\left(2sinx-1\right)\left(sinx-1\right)=0\)
⇔ \(\left(2sinx-1\right)\left[\sqrt{3}cos2x+2sinxcosx-2cosx\right]=0\)
⇔ \(\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\\sqrt{3}cos2x+sin2x=2cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x=cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(2x-\dfrac{\pi}{6}\right)=cosx\end{matrix}\right.\)
