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Ta có:
$\left\{\begin{array}{l}\sin^2(3x-\dfrac{\pi}{4})\ge0\\\cos^2(x+\dfrac{\pi}{3})\ge0\end{array}\right.\Rightarrow \sin^2(3x-\dfrac{\pi}{4})+\cos^2(x+\dfrac{\pi}{3})\ge0$
Dấu bằng xảy ra khi:
$\left\{\begin{array}{l}\sin^2(3x-\dfrac{\pi}{4})=0\\\cos^2(x+\dfrac{\pi}{3})=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sin(3x-\dfrac{\pi}{4})=0\\\cos(x+\dfrac{\pi}{3})=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}3x-\dfrac{\pi}{4}=k\pi&,k\in\mathbb{Z}\\x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+l\pi&,l\in\mathbb{Z}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=\dfrac{\pi}{12}+k\dfrac{\pi}{3}&,k\in\mathbb{Z}\\x=\dfrac{\pi}{6}+l\pi&,l\in\mathbb{Z}\end{array}\right.$
Suy ra phương trình vô nghiệm.
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