b. ĐKXĐ \(x\ne\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow4tanx+6=\dfrac{1}{cos^2x}\)
\(\Leftrightarrow4tanx+6=tan^2x+1\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)
e.
\(\Leftrightarrow sin^22x+sin2x.cos2x+1=0\)
Nhậ thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^22x\)
\(\Rightarrow tan^22x+tan2x+\left(1+tan^22x\right)=0\)
\(\Leftrightarrow2tan^22x+tan2x+1=0\) (vô nghiệm)
Vậy pt đã cho vô nghiệm
h.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos4x\)
\(\Leftrightarrow1-\dfrac{3}{8}\left(1-cos4x\right)=cos4x\)
\(\Leftrightarrow cos4x=1\)
\(\Leftrightarrow4x=k2\pi\)
\(\Leftrightarrow x=\dfrac{k\pi}{2}\)
