<s>[INST] $ 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.
 [/INST] 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 $2 \pi$.
</s>