Ta có $\begin{cases} \sin^9 x \leqslant 1 \\ \cos^9 x \leqslant 1 \end{cases} \implies \begin{cases} \sin^{11} x \leqslant \sin^2 x \\ \cos^{11} x \leqslant \cos^2 x \end{cases}$
Cộng vế $\implies \sin^{11} x + \cos^{11} x \leqslant \sin^2 x + \cos^2 x = 1$
Vậy dấu '=' xảy ra tức $\sin^{11} x + \cos^{11} x = 1 \iff \begin{cases} \sin^{11} x = \sin^2 x \\ \cos^{11} x = \cos^2 x \end{cases}$
$\iff \begin{cases} \sin x = 0 \\ \cos^{11} x = \cos^2 x \end{cases} \vee \begin{cases} \sin x = 1 \\ \cos^{11} x = \cos^2 x \end{cases}$
$\iff \begin{cases} \sin x = 0 \\ \cos^{11} x = 1 \end{cases} \vee \begin{cases} \sin x = 1 \\ 0 = 0 \end{cases}$ (từ $\sin x = 0 \implies \cos^2 x = 1$, từ $\sin x = 1 \implies \cos x = 0$)
$\iff \begin{cases} x = k\pi \\ x = m2\pi \end{cases} \vee x = \dfrac{\pi}2 + k2\pi$
$\iff x = k2\pi \vee x = \dfrac{\pi}2 + k2\pi$ ($k \in \mathbb{Z}$)