Let  $f\left(x\right)=a{x}^2+bx+c,$  where a,b,c  $\in$ R and  $a\ne 0$. If  $a^2+c^2-b^2+2ac<0$  and exactly one root of 

f(x) = 0 lies in $(\alpha ,\beta )$  and  $\left|\alpha \right|$  and  $\left|\beta \right|$  are minimum, The number of integral values of n for which the equation  $\sin x(\sin x+\cos x)=\frac{n}{4}$  has at least one solution is

We have,
 $\Rightarrow a^2+c^2-b^2+2ac<0\Rightarrow {(a+c)}^2-b^2<0$
  $\Rightarrow (a+b+c)(a+c-b)<0\Rightarrow f\left(1\right)f(-1)<0$
So the equation f (x) =0 will have 2 distinct solutions as exactly one root lie between (-1,1)
 $\sin x(\sin x+\cos x)=\frac{n}{4}\Rightarrow 2{\sin }^{2}x+2\sin x\cos x=\frac{n}{2}$
 $1-\cos 2x+\sin 2x=\frac{n}{2}\Rightarrow \sin 2x-\cos 2x=\frac{n}{2}-1$
For the solution to be exists
 $-\sqrt{2}\le \frac{n}{2}-1\le \sqrt{2}\Rightarrow 2(1-\sqrt{2})\le n\le 2(\sqrt{2}+1)$
 $\Rightarrow -0.828\le n\le 4.828$
 $\therefore$  Number of integral values of n is 5