Question

Sin³x - cos³x = 1 + sinxcosx

Answer 1

$\sin^3x-\cos^3x=1+\sin x\cos x\\ (\sin x-\cos x)(1+\sin x\cos x)-(1+\sin x\cos x)=0\\ (1+\sin x\cos x)(\sin x-\cos x-1)=0\\ 1+\sin x\cos x=0\\ \sin^2x+\cos^2x+\sin x\cos x=0|:\cos^2x\\ tg^2x+tgx+1=0\\ D=1^2-4\ \textless \ 0$
Уравнение решений не имеет
$\sin x-\cos x-1=0\\ \sin x-\cos x=1\\ \sqrt{2} \sin(x- \frac{\pi}{4})=1\\ \\ x- \frac{\pi}{4}=(-1)^k\,*\frac{\pi}{4}+ \pi k,k \in Z\\ x=(-1)^k\, *\frac{\pi}{4}+\frac{\pi}{4}+\pi k,k \in Z$