Question

решите уравнение
1-sin2x= -(sinx+cosx)
найдите все корни на промежутке -3П/2;П

Answer 1

$1-sin2x=-(sinx+cosx)\\t=sinx+cosx\\t^2=(sin^2x+cos^2x)+2sinxcosx=1+sin2x; ; to ; ; sin2x=t^2-1\\\1-(t^2-1)=-t\\t^2-t-2=0\\t_1=2,t_2=-1; (teor.; Vieta)\\1); sinx+cosx=2|:sqrt2$

$frac{1}{sqrt2}sinx+frac{1}{sqrt2}cosx=frac{2}{sqrt2}\\cosfrac{pi}{4}cdot sinx+sinfrac{pi}{4}cdot cosx=frac{2}{sqrt2}\\sin(x+frac{pi}{4})=frac{2}{sqrt2}>1; to ; net; reshenij\\2); sinx+cosx=-1|:sqrt2\\frac{1}{sqrt2}sinx+frac{1}{sqrt2}cosx=-frac{1}{sqrt2}\\sin(x+frac{pi}{4})=-frac{1}{sqrt2}\\x+frac{pi}{4}=(-1)^{n+1}cdot frac{pi}{4}+pi n,; nin Z\\x=(-1)^{n+1}cdot frac{pi}{4}-frac{pi}{4}+pi n$

$x= left { {{pi n,esli; n-nechetnoe} atop {-frac{pi}{2}+pi n,esli; n- chetnoe}} right.$