5 Obtain the two integer values of $x$ closest to $2013^{\circ}$, both by default and by excess, that satisfy this trigonometric equation

$$
2^{\sin ^{2} x}+2^{\cos ^{2} x}=2 \sqrt{2}
$$
Solution. Applying the arithmetic and geometric mean inequality results in

$$
2^{\sin ^{2} x}+2^{\cos ^{2} x} \geq 2 \sqrt{2^{\sin ^{2} x} \cdot 2^{\cos ^{2} x}}=2 \sqrt{2^{\sin ^{2} x+\cos ^{2} x}}=2 \sqrt{2}
$$

Equality is achieved when $2^{\sin ^{2} x}=2^{\cos ^{2} x}$. That is, when $\sin ^{2} x=\cos ^{2} x$ or $\sin x= \pm \cos x$. The values of $x$ that satisfy the above equality are $x=45^{\circ}+90^{\circ} k$ with $k \in \mathbb{Z}$. The required values are obtained for

$$
k_{1}=\left[\frac{2013^{\circ}-45^{\circ}}{90}\right]=21 \quad \text { and } \quad k_{1}=\left[\frac{2013^{\circ}+45^{\circ}}{90}\right]=22
$$

and are $x_{1}=1935^{\circ}$ and $x_{2}=2025^{\circ}$.