Question

Find general solution of the equation ${\sin }^{3}x-{\cos }^{3}x=1+\sin x\cos x$

Answer 1

${\sin }^{3}x-{\cos }^{3}x=1+\sin x\cos x$

$(\sin x-\cos x)({\sin }^{2}x+{\cos }^{2}x+\sin x\cos x)=1+\sin x\cos x$

$(\sin x-\cos x)(1+\sin x\cos x)=1+\sin x\cos x$

$(\sin x-\cos x)=1$

$\sin x-\cos x-1=0$

$-\sqrt{2}\cos (\frac{\pi }{4}+x)-1=0$

$-\sqrt{2}\cos (\frac{\pi }{4}+x)=1$

$\cos (\frac{\pi }{4}+x)=-\frac{\sqrt{2}}{2}$

$\frac{\pi }{4}+x=\frac{3\pi }{4}+2n\pi ,\frac{\pi }{4}+x=\frac{5\pi }{4}+2n\pi$

$\Rightarrow x=\frac{\pi }{2}+2\pi n,x=\pi +2\pi n$