Problem 1. Solve the equation: $2^{\sin ^{4} x-\cos ^{2} x}-2^{\cos ^{4} x-\sin ^{2} x}=\cos 2 x$.
Solution. The equation is equivalently written as $2^{\sin ^{4} x+\sin ^{2} x}-2^{\cos ^{4} x+\cos ^{2} x}=2 \cos 2 x$. Let's observe that

$$
\begin{aligned}
\cos ^{4} x+\cos ^{2} x-\sin ^{4} x-\sin ^{2} x & =\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)+\cos ^{2} x-\sin ^{2} x \\
& =2 \cos 2 x
\end{aligned}
$$

and the equation becomes $2^{\sin ^{4} x+\sin ^{2} x}+\sin ^{4} x+\sin ^{2} x=2^{\cos ^{4} x+\cos ^{2} x}+\cos ^{4} x+\cos ^{2} x$.

3 points

1 point

The function $f: \mathbb{R} \rightarrow \mathbb{R}, f(t)=2^{t}+t$ is strictly increasing, hence injective.

2 points

thus we obtain $\sin ^{4} x+\sin ^{2} x=\cos ^{4} x+\cos ^{2} x$, i.e., $2 \cos 2 x=0$, with solutions $x \in\left\{\left.\frac{(2 k+1) \pi}{4} \right\rvert\, k \in \mathbb{Z}\right\} \ldots 1$ point