Let [a] denote the greatest integer less than or equal to a. The set…

We have,

$1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2$

$=1+\frac{1}{2}\left(\frac{1-\sin^2θ}{1-\sin θ}\right)=1+\frac{1}{2}(1+\sin θ)=\frac{3}{2}+\frac{1}{2}\sin θ$

Now,

$-\frac{π}{2}≤θ\frac{π}{2}$

$⇒-1≤\sin θ<1$

$⇒-\frac{1}{2}≤\frac{1}{2}\sin θ<\frac{1}{2}$

$⇒\frac{3}{2}-\frac{1}{2}≤\frac{3}{2}+\frac{1}{2}\sin θ<\frac{3}{2}+\frac{1}{2}$

$⇒1≤1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2<2$

$⇒\left[1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2\right]=1$

Thus, the given inequality reduces to 

$|2x^2-4x-7|<1$

$⇒-1<2x^2-4x-7<1$

$⇒-1 <2x^2-4x-7$ and $2x^2 -4x-7 <1$

$⇒ 2x^2-4x-6>0$ and $2x^2-4x-8<0$

$⇒x^2 -2x-3> 0$ and $x^2 - 2x - 4 <0$

$⇒(x < -1\, or,\, x> 3)$ and $1-\sqrt{5} <x<1+ \sqrt{5}$

$⇒x∈(1-\sqrt{5},-1)∪(3,1+\sqrt{5})$