Question

РЕШИТЕ УРАВНЕНИЕ

4 + 2sin2x - 5sinx - 5cosx=0

Answer 1

$4+2sin2x-5sinx-5cosx=0\\4+2, sin2x-5(sinx+cosx)=0\\t=sinx+cosx; ,\\t^2=(sinx+cosx)^2=sin^2x+cos^2x+2, sinxcdot cosx=1+sin2x; ,\\sin2x=t^2-1\\4+2(t^2-1)-5t=0\\2t^2-5t+2=0; ,; ; D=9; ,; t_1=frac{1}{2}; ,; t_2=2\\a); ; sinx+cosx=frac{1}{2}; |:sqrt2\\frac{1}{sqrt2}, sinx+frac{1}{sqrt2}, cosx=frac{1}{2sqrt2}\\sinfrac{pi }{4}cdot sinx+cosfrac{pi }{4}cdot cosx=frac{sqrt2}{4}\\cos(x-frac{pi }{4})=frac{sqrt2}{4}\\x-frac{pi}{4}=pm arccosfrac{sqrt2}{4}+2pi n,; nin Z$

$x=frac{pi }{4}pm arccosfrac{sqrt2}{4}+2pi n,; nin Z\\b); ; sinx+cosx=2; |:sqrt2\\cos(x-frac{pi }{4})=sqrt2>1; ; Rightarrow ; ; xin varnothing ; ; (-1leq cosxleq 1)\\Otvet:; ; x=frac{pi }{4}pm arccosfrac{sqrt2}{4} +2pi n,; nin Z$