Seja $Z = \large{\frac{2\cdot \log_2 \sin(x)\space + \space 1}{i(e^{2ix}\space -2\cdot \cos^2(x) \space + \space 1)}}$ . Sabendo que $e^{2ix}$ é a forma exponencial do número complexo $\cos(2x) + i\sin(2x)$, então:$$e^{2ix} -2\cdot \cos^2(x) +1 = \cos^2x - \sin^2(x) + i\sin(2x) -2\cdot \cos^2(x) +1$$$$e^{2ix} -2\cdot \cos^2(x) +1 = i\sin(2x)- \underbrace{[\sin^2(x)+\cos^2(x)]}_{1} + 1$$Assim:$$e^{2ix} -2\cdot \cos^2(x) +1 = i\sin(2x)$$$$i(e^{2ix} -2\cdot \cos^2(x) +1) = -\sin(2x)$$Dessa forma, temos:$$Z = -\underbrace{\frac{2\cdot \log_2 \sin(x)+ 1}{\sin(2x)}}_{\mathbb{R}} > 0$$E isto implica que:$$\color{green}{\frac{2\cdot \log_2 \sin(x) + 1}{\sin(2x)} < 0}$$ I) Para $2\cdot \log_2 \sin(x) + 1 < 0 \implies \sin(x) < \frac{\sqrt 2}{2} = \sin(\frac{\pi}{4})$, temos: $x \in [0, \frac{\pi}{4})\cup(\frac{3\pi}{4}, 2\pi]$ II) Para $2\cdot \log_2 \sin(x) + 1 > 0 \implies \sin(x) > \frac{\sqrt 2}{2} = \sin(\frac{\pi}{4})$, temos: $x \in ( \frac{\pi}{4},\frac{3\pi}{4})$ III) Para $\sin(2x) < 0$, temos: $x \in (\frac{\pi}{2}, \pi)$ IV) Para $\sin(2x) > 0$, temos: $x \in (0, \frac{\pi}{2})$ Na inequação, resolve-se assim: $\text{I)} \space \cap \space \text{IV)} \implies [0, \frac{\pi}{4})\cup(\frac{3\pi}{4}, 2\pi] \space \cap \space (0, \frac{\pi}{2}) = \Large{(0, \frac{\pi}{4})}$ $\text{II)} \space \cap \space \text{III)} \implies ( \frac{\pi}{4},\frac{3\pi}{4}) \space \cap \space (\frac{\pi}{2}, \pi) = \Large{(\frac{\pi}{2}, \frac{3\pi}{4})}$ Portanto, o conjunto-solução $S$ é: $\space\boxed{S = \left(0, \frac{\pi}{4}\right)\cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right)}$ ou: $\boxed{S = \left\{ x\in \mathbb{R}\space | \space 2k\pi < x < \frac{\pi}{4} + 2k\pi \space \space \space \text{ou} \space \space \space 2k\pi + \frac{\pi}{2} < x < \frac{3\pi}{4} + 2k\pi, \space k \in \mathbb{Z} \right \}}$