$ 2 \cos ^{2021} x+\sin ^{2022} x=2 $

Question

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.

Answer 1

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 $2 \pi$.

Answer 2

Since you already received good answers and comments, use Taylor series around $x=0$ to $O(x^6)$ for $$f_n(x)=2 \cos ^{2 n+1}(x)+\sin ^{2 n+2}(x)-2$$ and look at the patterns for very small values of $n$.

They all write $$f_n(x)=-(2n+1)x^2+\frac {a_n}{12}\,x^4+O(x^6)$$ where the $a_n$ form the sequence $$\{33,65,133,225,341,481,645,833,1045,1281,1541,1825,2133,2465,2821,\cdots\}$$

So, suppose that you want to solve for $x$ $$2 \cos ^{2 n+1}(x)+\sin ^{2 n+2}(x)=2-\epsilon$$ just solve the quadratic in $x^2$.

For example, using $n=8$, $a_8=833$ and $\epsilon=\frac 1{10}$ would give $$x \sim \frac{1}{7} \sqrt{\frac{1}{85} \left(510-\sqrt{235110}\right)}=0.077659$$ while the exact solution (as given by Newton method) is $x=0.077643$

Answer 3

Since $-1 \leq \sin x \leq 1$, $(\sin x)^{2022} \leq 1$, according to the given equation $(\cos x)^{2021}\gt 0$,therefore $0 \lt \cos x \lt 1$, which implies that $$(\cos x)^{2021} \leq \cos x \qquad {(1*)}$$.

$0 \leq (\sin x)^2 \leq 1$ gives $$(\sin x)^{2022} \leq (\sin x)^2 \qquad {(2*)}$$ $(1*) \times 2 + (2*)$ gives $$2( \cos x)^{2021} + (\sin x)^{2022} \leq 2\cos x + (\sin x)^2$$ $$2 \leq 2\cos x + (\sin x)^2$$ $$2 - 2\cos x \leq 1 - (\cos x)^2$$ $$2( 1- \cos x) \leq (1-\cos x)(1+\cos x)$$ $$2 \leq 1+ \cos x$$ $$1\leq \cos x$$ Since $\cos x$ can not be greater than $1$, $\cos x =1$, which gives $x = 2n\pi$ as the solution of $x$, where $n$ is an integer.