Question

Solve $1+{\sin }^{3}x+{\cos }^{3}x-\frac{3}{2}\sin 2x$.

Answer 1

${\sin }^{3}x+{\cos }^{3}x+1-3\sin x\cdot \cos x\cdot 1=0$
Above is of the form $a^3+b^3+c^3-3abc=0$
or $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
or $\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$\therefore a+b+c=0$ only as other factor is not zero.
or $\sin x+\cos x=-1$
$\cos (x-\frac{\pi }{4})=-\frac{1}{\sqrt{2}}=\cos \frac{3\pi }{4}$
$\therefore x-\frac{\pi }{4}=2n\pi \pm \frac{3\pi }{4}$
$\therefore x=2n\pi +\pi ,2n\pi -\frac{\pi }{2}$