Question

Cos^4x-sin^4x=cos9x Можно решение?)

Answer 1

   $cos^4x-sin^4x=cos9x\\ (1-sin^2x)^2-sin^4x=cos9x\\\ 1-2sin^2x=cos9x\\ cos2x=cos9x\\ cos2x-cos9x=0\\ sin(5.5x)*sin(3.5x)=0\\ \left \{ {{sin(5.5x)=0} \atop {sin(3.5x)=0}} \right. \\ x=\frac{2\pi*n}{11}\\ x=2\pi*n\\ x=2\pi*n-\frac{6\pi}{7}\\ x=2\pi*n-\frac{4\pi}{7}\\ x=2\pi*n-\frac{2\pi}{7}\\ x=2\pi*n+\frac{2\pi}{7}$