Question

(sin2x+√3 cos2x )^2=5+cos(π/6-2x)

Answer 1

$(sin2x+sqrt3cos2x)^2=5+cos( frac{pi}{6}-2x)\\\star ; ; sin2x+sqrt3cos2x=2cdot ( frac{1}{2}cdot sin2x+frac{sqrt3}{2}cdot cos2x)=\\=2cdot (sinfrac{pi}{6}cdot sin2x+cos frac{pi}{6}cdot cos2x)=2cdot cos( frac{pi}{6} -2x); ; star \\\4cos^2( frac{pi}{6}-2x) -cos( frac{pi}{6}-2x)-5=0 \\t=cos( frac{pi}{6} -2x); ,; -1 leq t leq 1; ,; ;5t^2-t-5=0; ,\\D=81; ,; ; t_1=frac{1-9}{8}=-1; ,; t_2=frac{10}{8}=frac{5}{4} textgreater 1\\cos( frac{pi}{6}-2x)= frac{1-9}{8}=-1$

$frac{pi}{6} -2x=-pi +2pi n,; nin Z\\2x=-frac{5pi}{6}+2pi n; ,; nin Zquad (period; mozno; pisat:; ; pm 2pi n)\\x=-frac{5pi}{12}+pi n,; nin Z$