Do argumento de $z$, temos: \begin{matrix} \tan{\dfrac{\pi}{4}} &=& 1 &=& { \dfrac{ \left(\dfrac{1-2\cos{\alpha}+2\sin{\alpha}}{\sin{2\alpha}}\right)} {\left(\dfrac{1-\cos{\alpha}}{\sin{\alpha}\cos{\alpha}}\right)\cdot \color{royalblue}{\dfrac{2}{2}} } } &=& \Large{ \frac{ (1-2\cos{\alpha}+2\sin{\alpha})}{2(1-\cos{\alpha})}} \end{matrix} Continuando, \begin{matrix} 2(1-\cos{\alpha}) = 1-2\cos{\alpha}+2\sin{\alpha} &\Rightarrow& \sin{\alpha} = {\dfrac{1}{2}} \end{matrix} Assim, segundo o intervalo de $\alpha$ fornecido pelo enunciado, concluímos que: \begin{matrix} \fbox{$\alpha = {\dfrac{\pi}{6}}$} \\ \\ Letra \ (A) \end{matrix} $\color{orangered}{Obs:}$ $\color{}{ \sin{2x} = 2\sin{x}\cos{x} }$