1.
\(4sin^22x-3sin2x.cos2x-cos^22x=0\)
Với \(cos2x=0\) không phải nghiệm
Với \(cos2x\ne0\), chia 2 vế cho \(cos^22x\)
\(4tan^22x-3tan2x-1=0\)
\(\Leftrightarrow\left(tan2x-1\right)\left(4tan2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tan2x=1\\tan2x=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{4}+k\pi\\2x=arctan\left(-\dfrac{1}{4}\right)+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{1}{2}arctan\left(-\dfrac{1}{4}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9\pi}{8}\\x=\dfrac{13\pi}{8}\\x=\dfrac{1}{2}arctan\left(-\dfrac{1}{4}\right)+\dfrac{3\pi}{2}\\x=\dfrac{1}{2}arctan\left(-\dfrac{1}{4}\right)+2\pi\end{matrix}\right.\)
2.
\(4sin^22x-1=0\)
\(\Leftrightarrow2\left(1-cos4x\right)-1=0\)
\(\Leftrightarrow cos4x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{3}+k2\pi\\4x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}\\x=\dfrac{5\pi}{12}\end{matrix}\right.\)
