Question

Решите два уравнения:
1) √3sin²x - 4sinxcosx + √3cos²x = 0
2) sinxcosx = -0.25

Answer 1

$1)quad sqrt3sin^2x-4sinxcdot cosx+sqrt3cos^2x=0; |:cos^2xne 0\\sqrt3tg^2x-4tgx+sqrt3=0\\t=tgx; ,; ; sqrt3t^2-4t+sqrt3=0\\D/4=4-sqrt3cdot sqrt3=4-3=1\\t_1=frac{2-1}{sqrt3}= frac{1}{sqrt3} =frac{sqrt3}{3}; ;; ; t_2= frac{2+1}{sqrt3} = frac{3}{sqrt3} =sqrt3\\a)quad tgx=frac{sqrt3}{3}; ,; ; x=frac{pi}{6}+pi n,; nin Z\\b)quad tgx=sqrt3; ,; ; x=frac{pi}{3}+pi k,; kin Z$

$2)quad sinx cdot c osx=-0,25\\frac{1}{2}sin2x=-frac{1}{4}\\sin2x=-frac{1}{2}\\2x=(-1)^{n}arcsin(-frac{1}{2})+pi n\\2x=(-1)^{n}(-frac{pi}{6})+pi n=(-1)^{n+1}frac{pi}{6}+pi n\\x=(-1)^{n+1}frac{pi}{12}+frac{pi n}{2}; ,; nin Z$