Question

тригонометрические уравнения..... .

Answer 1

$1); ; 6cos^4x-11cos^2x+4=0\\t=cos^2x; ,; ; 0 leq cos^2x leq 1; ; to ; 0 leq tleq 1\\6t^2-11t+4=0; ,; ; D=25; ,\\t_1= frac{11-5}{12}=frac{1}{2} textless 1; ,; ; t_2= frac{11+5}{12}= frac{4}{3} textgreater 1\\cos^2x= frac{1}{2}; ; Rightarrow ; ; frac{1+cos2x}{2} = frac{1}{2} ; ; Rightarrow ; ; cos2x=0\\2x=frac{pi}{2}+pi n; ,; ; ; x=frac{pi}{4}+frac{pi n}{2}; ,; nin Z\\xin [-frac{7pi}{2};-2pi ]:; ; x=- frac{9pi }{4} ; ,; -frac{11pi }{4}; ,; ; -frac{13pi }{4}; .$

$2); ; (2cos^2x-cosx)sqrt{-11tgx}=0; ,; ; ODZ:; ; xne frac{pi }{2}+pi n,; nin Z\\ left { {{cosx(2cosx-1)=0} atop {tgx leq 0; ,; xne frac{pi}{2}+pi n}} right. ; ; left { {{cosx=0; ; ili; ; cosx=frac{1}{2}} atop {-frac{pi}{2}+pi n textless x leq pi n; ,; xne frac{pi}{2}+pi n}} right. \\ left { {{x=pm frac{pi}{3}+2pi n} atop {-frac{pi}{2}+pi n textless xleq pi n}} right. ; ; to ; ; x=-frac{pi}{3}+2pi n; ,; ; x=2pi n; ,; nin Z\\Otvet:; ; x=- frac{pi }{3}+2pi n; ,; ; x=2pi n; ,; nin Z; .$