Sin 2x-5sin x+5cos x+5=0

$sin2x-5sin x+5cos x+5=0 sin2x-5(sin x-cos x)+5cos^2x+5sin^2x=0 sin2x -5(sin x-cos x)+10sin 2x+5(sin x-cos x)^2=0 5(sin x-cos x)^2-5(sin x-cos x)+11sin 2x=0$
Пусть $sin x-cos x=t$ $(|t| leq sqrt{2} )$ , тогда $(sin x-cos x)^2=t^2$ . откуда
$1-sin2x=t^2 sin2x = 1-t^2$

Заменяем
$11(1-t^2)-5t+5t^2=0 11-11t^2-5t+5t^2=0 6t^2+5t-11=0$
Находим дискриминант
$D=b^2-4ac=25+264=289; sqrt{D} =17$
$t_1= frac{-5-17}{12}=- frac{11}{6} ,,notin|t| leq sqrt{2}$
$t_2= frac{-5+17}{12} =1$

Возвращаемся к замене
$sin x-cos x=1 sqrt{2} sin(x- frac{pi}{4} )=1 sin(x- frac{pi}{4})= frac{1}{ sqrt{2} } x- frac{pi}{4}=(-1)^kcdot frac{pi}{4}+pi k,k in Z x=(-1)^kcdot frac{pi}{4}+ frac{pi}{4}+pi k,k in Z$