$PT⇔\sin^2 4x+1-\sin^2 2x=0\\⇔\sin^2 4x+\cos^2 2x=0$
`0≤sin^2 4x≤1`
`0≤cos^2 2x≤1`
$→\sin^2 4x+\cos^2 2x≥0$
Dấu bằng xảy ra khi
$\begin{cases}\sin4x=0\\\cos2x=0\end{cases}\\⇔\begin{cases}x=\dfrac{kπ}{4}\\x=\dfrac{π}{4}+\dfrac{kπ}{2}\end{cases}\\→x=\dfrac{π}{4}+\dfrac{kπ}{2}(k\in\mathbb{Z})$