$ 2 \cos ^{2021} x+\sin ^{2022} x=2 $ Someone sent me this task, it is meant to be solved quickly, nothing to think too much about.
At first I tried representing $\sin ^{2022}x$ as $(\sin ^{2}x)^{1011}$
$$
2 \cdot \frac{\left(\cos ^{1011} x\right)^{2}}{\cos (x)}+\left(1-\cos ^{2} x\right)^{1011}=2.
$$
Then using using $t=\cos x = \sqrt{1-\sin ^{2}x}$ substitution and find $x$. The expression got too complex so I gave up the idea.
After some time considering Taylor's expansion, funtion series, etc I just realized the 'only' case when it satisfies the equality it's when $x=0$.
How would you solve it. How do you solve this equation analyticaly, I mean step by step, line by line.
There is no answer proposed in the book.

One elegant solution has been given in the comments. Here is an alternative one, using the Cauchy-Schwarz inequality. For any positive integer $n$ is
$$
\begin{align}
 f(x) &= 2 \cos^{2n+1}(x)+\sin^{2n+2}(x) \\
&= \cos^{2n}(x) \cdot 2 \cos(x) + \sin^{2n}(x) \cdot \sin^2(x) \\
&\le \sqrt{\cos^{4n}(x) + \sin^{4n}(x)} \cdot \sqrt{4 \cos^2(x) + \sin^4(x)} \, .
\end{align}
$$
Now
$$
 0 \le \cos^{4n}(x) + \sin^{4n}(x) \le \cos^{2}(x) + \sin^{2}(x) = 1
$$
and
$$
0 \le 4 \cos^2(x) + \sin^4(x) = 4 \cos^2(x) + (1 - \cos^2(x))^2
= (1+\cos^2(x))^2 \le 4 
$$
so that $f(x) \le 2$ for all $x\in \Bbb R$.
Equality holds if and only if $\cos^2(x) = 1$ and $\cos(x) \ge 0$, that is exactly for all integer multiples of $2 \pi$.
