Question

ПОМОГИТЕ
3sin2x+4cos2x=5

Answer 1

$3sin2x+4cos2x=5\\frac{sqrt3}{5}}sin2x+frac{2}{5}cos2x=1\\cos alpha =frac{sqrt3}{5}; ,; ; sin alpha =frac{2}{5}; ; to ; ; tg alpha =frac{2}{sqrt3}; ; to ; ; alpha =arctgfrac{2}{sqrt3}\\sin(2x+ alpha )=1\\2x+ alpha =frac{pi}{2}+2pi n; ,; nin Z\\2x=frac{pi}{2}- alpha +2pi n=frac{pi}{2}-arctgfrac{2}{sqrt3}+2pi n; ,; nin Z\\x=frac{pi}{4}-frac{1}{2}arctgfrac{2}{sqrt3}+pi n; ,; nin Z$