Question

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

Answer 1

$\sin2x-5\sin x+5\cos x+5=0\\ \sin2x-5(\sin x-\cos x)+5\cos^2x+5\sin^2x=0\\ \sin2x -5(\sin x-\cos x)+10\sin 2x+5(\sin x-\cos x)^2=0\\ 5(\sin x-\cos x)^2-5(\sin x-\cos x)+11\sin 2x=0$
  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
$\sin x-\cos x=1\\ \sqrt{2} \sin(x- \frac{\pi}{4} )=1\\ \sin(x- \frac{\pi}{4})= \frac{1}{ \sqrt{2} } \\x- \frac{\pi}{4}=(-1)^k\cdot \frac{\pi}{4}+\pi k,k \in Z\\ x=(-1)^k\cdot \frac{\pi}{4}+ \frac{\pi}{4}+\pi k,k \in Z$