Solve for $\displaystyle x$, $\displaystyle y$: $\displaystyle 25^{\sin^2x+2\cos y +1}+25^{\cos^2y+2\sin x +1}=2$.
One possibility is that both those powers of 25 be equal to 1 so that 1+ 1= 2. In that case we must have $\displaystyle sin^2x+ 2 cos y+ 1= 0$ and $\displaystyle cos^2y+ 2\sin x+ 1= 0$. Adding those 2 equations, 1+ 2(cos x+ cos y)+ 2= 0 or cos x+ cos y= -3/2. From that cos y= -3/2- cos x. You can put that into $\displaystyle cos^2 x+ cos^2 y= 1$ to get a quadratic equation for cos x and then find x and y.
$\displaystyle \displaystyle 25^{\sin^2x+2\cos y+1}+25^{\cos^2y+2\sin x+1}\geq 2\sqrt{25^{(\sin x+1)^2+(\cos x+1)^2}}\geq 2$
The equality stands if $\displaystyle 25^{\sin^2x+2\cos y+1}=25^{\cos^2y+2\sin x+1}=1\Rightarrow $
$\displaystyle \left\{\begin{array}{ll}\sin^2x+2\cos y+1=0\\\cos^2y+2\sin x+1=0\end{array}\right.$
Adding the equations we have
$\displaystyle (\sin x+1)^2+(\cos y+1)^2=0\Rightarrow \sin x=\cos y=-1$
Now find x and y.