Given determinant is 

$\Delta =\left|\begin{array}{ccc}1+{\text{sin}}^{2}x& {\text{cos}}^{2}x& \text{sin 2x}\\ {\text{sin}}^{2}x& 1+{\text{cos}}^2\text{x}& \text{sin 2x}\\ {\text{sin}}^2\text{x}& {\text{cos}}^2\text{x}& 1+\text{sin 2x}\end{array}\right|$

Applying the operation ${\text{C}}_1\to {\text{C}}_1+{\text{C}}_2$, we get

$\Delta =\left|\begin{array}{ccc}2& {\text{cos}}^{2}x& \text{sin 2x}\\ 2& 1+{\text{cos}}^{2}x& \text{sin 2x}\\ 1& {\text{cos}}^{2}x& 1+\text{sin 2x}\end{array}\right|$

Applying the operations  ${\text{R}}_2\to {\text{R}}_2-{\text{R}}_1$  and then ${\text{R}}_3\to {\text{R}}_3-{\text{R}}_1$, we get

$\Delta =\left|\begin{array}{ccc}2& {\text{cos}}^{2}x& \text{sin 2x}\\ 0& 1& 0\\ -1& 0& 1\end{array}\right|$

Now, expanding the above determinant along $R_2$, we get

$\Delta =-0+(2+\text{sin 2x})-0=2+\text{sin 2x}$

Since, the maximum value of $\text{sin}\text{2x}$ is $1$ and minimum value of $\text{sin}\text{2x}$ is $-1$.

Therefore, $\alpha =$maximum value of $\Delta =2+1=3$  and  $\beta =$ minimum value of $\Delta =2-1=1$.

$\Rightarrow \alpha =3$  and $\beta =1$

$\therefore \alpha +2\beta =3+2=5$