Question

The number of integral value(s) of $k$ for which ${\sin }^{2}x+(\cos x-1)\sin x-\cos x-k\sin x+k=0$ have exactly three real roots, where $x\in (0,2\pi )$, is

Answer 1

Given equation is
${\sin }^{2}x+(\cos x-1)\sin x-\cos x-k\sin x+k=0\Rightarrow {\sin }^{2}x+\sin x\cdot \cos x-k\sin x-\sin x-\cos x+k=0\Rightarrow \sin x(\sin x+\cos x-k)-1(\sin x+\cos x-k)=0\Rightarrow (\sin x-1)(\sin x+\cos x-k)=0\Rightarrow \sin x=1\text{or}\sin x+\cos x=k$

As $x\in (0,2\pi )$, then $\sin x=1$ has one real root.
$\sin x+\cos x=k\Rightarrow \sin (x+\frac{\pi }{4})=\frac{k}{\sqrt{2}}$
This equation should have two real root, so
$k\in (-\sqrt{2},\sqrt{2})$
Therefore, the integral value of $k$ are $-1,0,1$
When $k=1$
$x=\frac{\pi }{2},(\because x\in (0,2\pi \left)\right)$

Therefore, $k=-1,0$
Hence, there are two possible integral values of $k.$