Question

6sin²x-5sinx-4=0
найти корни на отрезке [-7П/2; -3П/2]

Answer 1

$6sin^2x-5sinx-4=0; ,; ; ; [- frac{7pi }{2} ;- frac{3pi}{2} ]\\D=25+4cdot 6cdot 4=121\\(sinx)_1= frac{5-11}{12}=- frac{1}{2}quad iliquad (sinx)_2= frac{5+11}{12}= frac{4}{3} \\a); ; sinx=-frac{1}{2}; ; ; to ; ; ; x=(-1)^{n}cdot (-frac{pi}{6})+pi n; ,; nin Z \\x=(-1)^{n+1}cdot frac{pi}{6}+pi n; ,; nin Z\\b); ; sinx=frac{4}{3} textgreater 1; ; net; ; reshenij,; t.k.; ; |sinx| leq 1\\2)quad xin [-frac{7pi}{2};-frac{3pi}{2}]:\\x=-2pi -frac{pi}{6}=-frac{13pi }{6}; ; ,; ; ; x=-2pi -frac{5pi}{6}=-frac{17pi }{6}\\Otvet:; ; x=(-1)^{n+1}cdot frac{pi}{6}+pi n,; nin Z; ;; x=-frac{13pi }{6}; ,; x=-frac{17pi }{6}; .$