Question

помогите ,пожалуйста
sin2x+5(sinx+cosx)=0

Answer 1

$sin2x+5(sinx+cosx)=0\\t=sinx+cosx; ,\\ t^2=sin^2x+cos^2x+2sinxcdot cosx=1+sin2x,\\sin2x=t^2-1\\t^2+5t-1=0\\D=25+4=29; ,; t_{1,2}= frac{-5pm sqrt{29}}{2} \\t_1approx-5,19; ; ;; ; t_2approx0,19\\a)quad sinx+cosx= frac{-5-sqrt{29}}{2} , left |:sqrt2$

$frac{1}{sqrt2}sinx+frac{1}{sqrt2}cosx= frac{-5-sqrt{29}}{2sqrt2} \\cosfrac{pi}{4}cdot sinx+sinfrac{pi}{4}cdot cosx=frac{-5-sqrt{29}}{2sqrt2}\\frac{-5-sqrt{29}}{2sqrt2}approx -3,71$

$sin(frac{pi}{4}+x)approx -3,71 textless -1; ,; ; a; ; -1 leq sin(frac{pi}{4}+x) leq 1\\net; ; reshenij\\b)quad sinx+cosx=frac{-5+sqrt{29}}{2}|:sqrt2 \\sin(frac{pi}{4}+x)= frac{-5+sqrt{29}}{2sqrt2}}\\frac{pi}{4}+x=(-1)^{n}cdot arcsinfrac{-5+sqrt{29}}{2sqrt2}+pi n,; ; nin Z$

$x=-frac{pi}{4}+(-1)^{n}cdot arcsinfrac{-5+sqrt{29}}{2sqrt2}+pi n$