4sin2x+8(sinx-cosx)=7

$4\sin 2x+8(\sin x-\cos x)=7\\ 4\sin 2x+8(\sin x-\cos x)=7(\sin^2x+\cos^2x)\\ 4\sin2x+8(\sin x-\cos x)=7(\sin^2x+\cos^2x-\sin2x+\sin2x)=0\\ 4\sin2x+8(\sin x-\cos x)=7((\sin x - \cos x)^2+\sin 2x)\\ 4\sin 2x+8(\sin x-\cos x)=7(\sin x-\cos x)^2+7\sin 2x\\ 7(\sin x-\cos x)^2-8(\sin x-\cos x)+3\sin 2x=0$
Пусть $\sin x-\cos x=t\,\,(|t| \leq \sqrt{2} )$ , тогда
$(\sin x-\cos x)^2=t^2\\ 1-\sin2x=t^2\\ \sin2x = 1-t^2$
В результате замены переменных, получаем уравнение
$7t^2-8t+3(1-t^2)=0\\ 4t^2-8t+3=0\\ D=b^2-4ac= 64-48=46;\,\, \sqrt{D} =4\\ t_1= \frac{8-4}{8}=0.5$
$t_2= \frac{8+4}{8}=1.5\notin|t| \leq \sqrt{2}$

Возвращаемся к замене
$\sin x-\cos x=0.5\\ \sqrt{2}\sin(x- \frac{\pi}{4})=0.5|:\sqrt{2} \\ \sin (x- \frac{\pi}{4})= \frac{\sqrt{2}}{4} \\ x- \frac{\pi}{4}=(-1)^k\cdot \arcsin(\frac{\sqrt{2}}{4})+ \pi k,k \in Z \\ x=(-1)^k\cdot \arcsin(\frac{\sqrt{2}}{4})+\frac{\pi}{4}+ \pi k,k \in Z$