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Ta có:
$\left\{\begin{array}{l}\sin^62x+1\ge 1\\\sin^2x\le1\end{array}\right.\Rightarrow \sin^62x+1\ge\sin^2x$
Dấu bằng xảy ra khi:
$\left\{\begin{array}{l}\sin2x=0\\\sin^2x=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sin2x=0\\\cos x=0\end{array}\right.$
$\Leftrightarrow \cos x=0$
$\Leftrightarrow x=\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$
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