Question

Помогите, пожалуйста, решить уравнения.

Answer 1

$1); ; sin^2x(sinx+3cosx)+cos^2x(cosx+3sinx)+2sinx(1-3cosx)+\\+2cosx(1-3sinx)=3\\underbrace {sin^3x+3sin^2x, cosx+cos^3x+3sinx, cos^2x}_{(sinx+cosx)^3}+underline {2sinx}-6sinx,cosx+underline {2cosx}-\\-6sinx, cosx=3\\(sinx+cosx)^3+2(sinx+cosx)-12sinx, cosx=3$

$star ; ; ; t=sinx+cosx; ,\\ t^2=(sinx+cosx)^2=underbrace {sin^2x+cos^2x}_{1}+2sinx, cosx=1+2sinx, cosx\\Rightarrow ; ; ; sinx, cosx=frac{t^2-1}{2}; ; star \\t^3+2t-6(t^2-1)=3\\t^3-6t^2+2t+3=0; ,; ; (t=1; ; -; ; koren)$

$t^3-6t^2+2t+3=(t-1)(t^2-5t-3)=0\\a); ; t_1=1; ; Rightarrow ; ; sinx+cosx=1; |:sqrt2\\ frac{1}{sqrt2}cdot sinx+ frac{1}{sqrt2}cdot cosx=frac{1}{sqrt2}; ; ,; ; frac{1}{sqrt2}=frac{sqrt2}{2} \\sinfrac{pi}{4}cdot sinx+cosfrac{pi}{4}cdot cosx=frac{sqrt2}{2}\\cos(x-frac{pi}{4})=frac{sqrt2}{2}\\x-frac{pi}{4}=pm frac{pi}{4}+2pi n,; nin Z\\boxed {x= frac{pi}{4}pm frac{pi}{4}+2pi n}= left [ {{2pi n,; nin Z} atop {frac{pi}{2}+2pi n,; nin Z}} right.$

$b); ; t^2-5t-3=0\\D=25+4cdot 3=37\\t_2= frac{5-sqrt{37}}{2}approx -0,541\\sinx+cosx= frac{5-sqrt{37}}{2}; |:sqrt2\\cos(x-frac{pi}{4})= frac{5-sqrt{37}}{2sqrt2} \\x-frac{pi}{4}=pm arccos frac{5-sqrt{37}}{2sqrt2}+2pi k,; kin Z\\boxed {x= frac{pi }{4} pm arccosfrac{5-sqrt{37}}{2sqrt2}+2pi k; ,; kin Z}$

$t_3=frac{5+sqrt{37}}{2}approx 5,541\\sinx+cosx= frac{5+sqrt{37}}{2} ; ; ; Rightarrow ; ; ; cos(x-frac{pi}{4})= frac{5+sqrt{37}}{2sqrt2} textgreater 1\\net; reshenij; ,; t.k.; ; |cosalpha |leq 1$

$Otvet:; ; x=frac{pi}{4}pm frac{pi}{4}+2pi n,; nin Z; ; ;\\x=frac{pi}{4}pm arccosfrac{5-sqrt{37}}{2sqrt2}+2pi k,; kin Z$

$2); ; 3cos^32x+11sin^22x-frac{4(1-tg^2x)}{1+tg^2x}=11\\star ; ; frac{1-tg^2x}{1+tg^2x}=frac{1-frac{sin^2x}{cos^2x}}{1+frac{sin^2x}{cos^2x}} =frac{cos^2x-sin^2x}{sin^2x+cos^2x} = frac{cos2x}{1} =cos2x; ; star \\3cdot cos^32x+11cdot underbrace {sin^22x}_{1-cos^22x}-4cdot cos2x-11=0\\3, cos^32x-11, cos^22x-4, cos2x=0\\t=cos2x; ,; ; ; 3t^3-11t^2-4t=0\\tcdot (3t^2-11t-4)=0$

$a); ; t_1=0; ; Rightarrow ; ; cos2x=0; ,; 2x=frac{pi}{2}+pi n,; nin Z\\boxed {x=frac{pi}{4}+frac{pi n}{2}; ,; nin Z}$

$b); ; 3t^2-11t-4=0; ,; ; ; D=121+48=169; ,\\t_1= frac{11-13}{6}=-frac{1}{3} ; ; to ; ; ; cos2x=-frac{1}{3}\\2x=pm arccos(-frac{1}{3})+2pi k,; kin Z\\2x=pm (pi -arccosfrac{1}{3})+2pi k,; kin Z\\boxed {x=pm frac{1}{2}cdot (pi -arccos frac{1}{3})+pi k,; kin Z }\\t_2= frac{11+13}{6}=4; ; to ; ; ; cosx2x=4 textgreater 1; -; ; net; reshenij \\Otvet:; ; x=frac{pi}{4}+frac{pi n}{2}; ,; nin Z; ;; ; x=pm frac{1}{2}cdot (pi -arccosfrac{1}{3})+pi k,; kin Z; .$