Trong khoảng \(\left( {0\,\,;\,\,\dfrac{\pi }{2}} \right)\) phương trình \({\sin ^2}4x + 3\sin 4x\cos 4x - 4{\cos ^2}4x = 0\) có:

Trường hợp 1: \(\cos 4x = 0 \Leftrightarrow 4x = \dfrac{\pi }{2} + k\pi \Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\,\,\left( {k \in Z} \right)\). Khi đó \({\sin ^2}4x = 1\)
Thay vào phương trình ta có: \(1 + 3.0 - 4.0 = 0 \Leftrightarrow 1 = 0\,\,\left( {Vô lý} \right)\)
\( \Rightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\,\,\left( {k \in Z} \right)\) không là nghiệm của phương trình.
Trường hợp 2: \(\cos 4x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\,\,\left( {k \in Z} \right)\). Chia cả 2 vế của phương trình cho \({\cos ^2}4x\) ta được:
\(\dfrac{{{{\sin }^2}4x}}{{{{\cos }^2}4x}} + 3\dfrac{{\sin 4x}}{{\cos 4x}} - 4 = 0 \Leftrightarrow {\tan ^2}4x + 3\tan 4x - 4 = 0\)
Đặt \(\tan 4x = t\). Khi đó phương trình trở thành
\({t^2} + 3t - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 1\\t = - 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\tan 4x = 1\\\tan 4x = - 4\end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l}4x = \dfrac{\pi }{4} + k\pi \\4x = \arctan \left( { - 4} \right) + k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}\\x = \dfrac{1}{4}\arctan \left( { - 4} \right) + \dfrac{{k\pi }}{4}\end{array} \right.\,\,\left( {k \in Z} \right)\)
Xét nghiệm \(x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}\,\,\left( {k \in Z} \right),\,x \in \left( {0;\dfrac{\pi }{2}} \right)\)
\( \Leftrightarrow \left\{ \begin{array}{l}0 < \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4} < \dfrac{\pi }{2}\\k \in Z\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}0 < \dfrac{1}{{16}} + \dfrac{k}{4} < \dfrac{1}{2}\\k \in Z\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l} - \dfrac{1}{4} < k < \dfrac{7}{4}\\k \in Z\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = 0\\k = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{\pi }}{{16}}\\x = \dfrac{{5\pi }}{{16}}\end{array} \right.\)
Xét nghiệm \(x = \dfrac{1}{4}\arctan \left( { - 4} \right) + \dfrac{{k\pi }}{4}\,\,\left( {k \in Z} \right);\,\,x \in \left( {0;\dfrac{\pi }{2}} \right)\)
\(\begin{array}{l}\left\{ \begin{array}{l}0 < \dfrac{1}{4}\arctan \left( { - 4} \right) + \dfrac{{k\pi }}{4} < \dfrac{\pi }{2}\\k \in Z\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l} - \dfrac{1}{4}\arctan \left( { - 4} \right) < \dfrac{{k\pi }}{4} < \dfrac{\pi }{2} - \dfrac{1}{4}\arctan \left( { - 4} \right)\\k \in Z\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}0,42 < k < 2,42\\k \in Z\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = 1\\k = 2\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{1}{4}\arctan \left( { - 4} \right) + \dfrac{\pi }{4}\\x = \dfrac{1}{4}\arctan \left( { - 4} \right) + \dfrac{\pi }{2}\end{array} \right.\end{array}\)
Vậy phương trình có 4 nghiệm thuộc khoảng \(\left( {0\,\,;\,\,\dfrac{\pi }{2}} \right)\)