Question

очень срочно нужно желательно на листочке

Answer 1

$1); ; sinx-cosx=1+sinxcdot cosx\\star ; t=sinx-cosx; ,; ; t^2=underline {sin^2x}-2sinxcdot cosx+underline {cos^2x}=1-2sinxcdot cosx; to \\2sinxcdot cosx=1-t^2; ,; ; sinxcdot cosx=frac{1-t^2}{2}; ; star \\t=1+frac{1-t^2}{2}; ; ,; ; t-1=frac{1-t^2}{2}; ; ,; ; 2(t-1)=1-t^2; ,; ; 2t-2=1-t^2; ,\\t^2+2t-3=0; ,; ; t_1=1; ,; t_2=-3; ; (teorema; Vieta)\\a); ; sinx-cosx=1; |:sqrt2\\frac{1}{sqrt2}cdot sinx-frac{1}{sqrt2}cdot cosx=frac{1}{sqrt2}$

$(frac{1}{sqrt2})^2+(frac{1}{sqrt2})^2=1; ; to ; ; sinfrac{pi}{4} =frac{1}{sqrt2}; ,; cosfrac{pi}{4}=frac{1}{sqrt2}; ; ; (sin^2a+cos^2a=1)\\cosfrac{pi}{4}cdot sinx-sinfrac{pi}{4}cdot cosx=frac{1}{sqrt2}\\sin(x-frac{pi}{4})=frac{1}{sqrt2}\\x-frac{pi }{4}=(-1)^{n}cdot frac{pi}{4}+pi n,; nin Z\\underline {x=frac{pi }{4}+(-1)^{n}cdot frac{pi }{4}+pi n; ,; nin Z}\\b); ; sinx-cosx=-3; |:sqrt2\\cosfrac{pi}{4}cdot sinx-sinfrac{pi }{4}cdot cosx=-frac{3}{sqrt2}$

$sin(x-frac{pi }{4})=-frac{3}{sqrt2}; ; ,; ; ; ; -frac{3}{sqrt2}<-1; ; ,; ; ; -1leq sin(x-frac{pi}{4})leq 1; ; Rightarrow \\xin varnothing \\Otvet:; ; x=frac{pi}{4}+(-1)^{n}cdot frac{pi}{4}+pi n; ,; nin Z; .$

$2); ; 2sin^2x+5sinxcdot cosx-5cos^2x=1\\2sin^2x+5sinxcdot cosx-5cos^2x=sin^2x+cos^2x\\sin^2x+5sinxcdot cosx-6cos^2x=0; |:cos^2xne 0\\tg^2x+5tgx-6=0; ,; D=49; ,; ; (tgx)_1=-6; ,; (tgx)_2=1\\a); tgx=-6; ,; underline {x=-arctg6+pi n,; nin Z}\\b); tgx=1; ,; underline {x=frac{pi}{4}+pi k,; kin Z}$