Question

Найдите сумму корней уравнения cosx-cos^2x-sin^3x=0, лежащих на отрезке [180*;540*]. Ответ запишите в градусах.

Answer 1

$cosx-cos^2x-sin^3x=0; ; ,; ; ; xin [, 180^circ;540^circ]\\cosx(1-cosx)-sin^2xcdot sinx=0\\cosx(1-cosx)-(1-cos^2x)cdot sinx=0\\cosx(1-cosx)-(1-cosx)(1+cosx)sinx=0\\(1-cosx)cdot (cosx-(1+cosx)sinx)=0\\(1-cosx)cdot (cosx-sinx-sinx, cosx)=0\\a); ; 1-cosx=0; ,; ; cosx=1; ,; ; x=2pi k,; kin Z\\xin [180^circ ;540^circ ]:; ; x_1=180^circ\\b); ; cosx-sinx-sinx, cosx=0; ,\\t=cosx-sinx; ,; t^2=cos^2x+sin^2x-2sinx, cosx=1-2sinx, cosx\\2sinx, cosx=1-t^2; ,; ; sinx, cosx=frac{1-t^2}{2}; .$

$t- frac{1-t^2}{2}=0; ,; ; t^2+2t-1=0 ; ,; ; frac{D}{4}=1^2-1cdot (-1)=1+1=2\\t_{1,2}=-1pm sqrt2\\t_1=-1-sqrt2; ,; ; t_2=-1+sqrt2\\a); ; cosx-sinx=-1-sqrt2; |:sqrt2\\ underbrace {frac{1}{sqrt2}}_{sinfrac{pi}{4}}cdot cosx- underbrace {frac{1}{sqrt2}}_{cosfrac{pi}{4}}cdot sinx= frac{-1-sqrt2}{sqrt2} \\sin(frac{pi}{4}-x)=-frac{1+sqrt2}{2} textless -1; ; (-frac{1+sqrt2}{sqrt2}approx -1,7 textless -1); ; net; reshenij,\\tak; kak; ; -1 leq sin(frac{pi}{4}-x) leq 1; .$

$b); ; cosx-sinx=frac{-1+sqrt2}{sqrt2}; ; ; ; (frac{-1+sqrt2}{sqrt2}approx 0,296)\\sin(frac{pi}{4}-x)=frac{sqrt2-1}{sqrt2}\\ frac{pi }{4}-x=(-1)^{n}cdot arcsinfrac{sqrt2-1}{sqrt2}+pi n= left [ {{arcsinfrac{sqrt2-1}{sqt2}+2pi n,; nin Z} atop {pi -arcsinfrac{sqrt2-1}{sqrt2}+2pi n,; nin Z}} right. \\x= left [ {{frac{pi}{4}-arcsinfrac{sqrt2-1}{sqrt2}-2pi n,; nin Z} atop {-frac{3pi}{4}+arcsinfrac{sqrt2-1}{sqrt2}-2pi n,; nin Z}} right.$

$c); ; xin [, 180^circ;540^circ]; ; ili; ; xin [pi ;3pi ]:\\x_2=frac{pi}{4}-arcsinfrac{sqrt2-1}{sqrt2}+2pi =frac{9pi}{4}-arcsinfrac{sqrt2-1}{sqrt2}; ;\\x_3=-frac{3pi}{4}+arcsinfrac{sqrt2-1}{sqrt2}+2pi =frac{5pi}{4}+arcsinfrac{sqrt2-1}{sqrt2}; ;\\x_2+x_3=frac{9pi}{4}+frac{5pi}{4}=frac{14pi}{4}=14cdot 45^circ=630^circ \\d); ; x_1+x_2+x_3=180^circ +630^circ=810^circ\\Otvet:; ; 810^circ ; .$