$-1<2{\sin }^2\theta <1$

I have tried the following: replace the $2{\sin }^2\theta$ with $1-\cos 2\theta$ then solve as normal...

$1-\cos 2\theta$

$\cos 2\theta <2$ - - - can't solve, right?

$1-\cos 2\theta <1$

$0<\cos 2\theta$

$\arccos (0)=\pi /2$

$\pi n\pm \pi /4<\theta$

I doubt that I have answered the question fully and sufficiently? What would the final inequality look like?

You've pretty much got it. Note that

Then you have

$0<\cos 2\theta$

Think of the cos graph here, and you'll see that $\cos x>0$ for $-\pi /2<x<\pi /2$ (and also adding $2\pi$ to $x$ will not change this, as this is the same as changing the position of the $y$-axis).

Therefore,

$2\theta =s+2n\pi$, where $\pi /2<s<\pi /2$ and $n$ is any integer

So, $\theta =t+n\pi$ with $\pi /4<t<\pi /4$
Regards,
Olof