Giải PT :
[tex]cos7x.cos5x-\sqrt{3}sin2x=1-sin7x.sin5x[/tex]
<=>[tex]cos7x.cos5x-\sqrt{3}.sin2x-1+sin7x.sin5x=(cos7x.cos5x+sin7x.sin5x)-\sqrt{3}.sin2x-1=cos(7x-5x)-\sqrt{3}.sin2x-1=(\frac{1}{2}.cos2x-\frac{\sqrt{3}}{2}.sin2x-\frac{1}{2}.cos2x).2=0[/tex]
<=>[tex]sin(2x-\frac{\pi }{6})=\frac{1}{2}[/tex]
<=>[tex]x=\frac{\pi }{6}+k\pi (k\in Z)[/tex];[tex]x=\frac{\pi }{2}+k\pi[/tex]