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[latex]2sin^2(\frac{8\pi }{2}+x)=\sqrt3cosx\\\\Formyla:\; sin^2 \alpha =\frac{1-cos2 \alpha }{2}\; \; \to \; \; 2sin^2 \alpha =1-cos2 \alpha \\\\1-cos(8\pi +2x)=\sqrt3cosx\\\\(1-cos2x)-\sqrt3cosx=0\\\\2sin^2x-\sqrt3cosx=0\\\\2(1-cos^2x)-\sqrt3cosx=0\\\\2cos^2x+\sqrt3cosx-2=0\\\\t=cosx,\; 2t^2+\sqrt3t-2=0[/latex] [latex]D=(\sqrt3)^2-4\cdot 2\cdot (-2)=3+16=19,\; \; \sqrt{19}\approx 4,36\\\\t_1=(cosx)_1=\frac{-\sqrt3-\sqrt{19}}{4}\approx -1,51<-1\; \; net\; reshenij\\\\t_2=(cosx)_2=\frac{-\sqrt3+\sqrt{19}}{4}\approx 0,66\\\\x=\pm arccos\frac{-\sqrt3+\sqrt{19}}{4}+2\pi n,\; n\in Z\\\\x=-arccos\frac{-\sqrt3+\sqrt{19}}{4}-2\pi\; \in [-\frac{7\pi }{2},-2\pi ][/latex]