1/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(6tan^2x+6tanx+2=\frac{1}{cos^2x}\)
\(\Leftrightarrow6tan^2x+6tanx+2=1+tan^2x\)
\(\Leftrightarrow5tan^2x+6tanx+1=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=-1\\tanx=-\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{1}{5}\right)+k\pi\end{matrix}\right.\)
b/
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow1-tanx-2tan^2x-\frac{1}{cos^2x}=0\)
\(\Leftrightarrow1-tanx-2tan^2x-1-tan^2x=0\)
\(\Leftrightarrow3tan^2x+tanx=0\)
\(\Leftrightarrow tanx\left(3tanx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=-\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=arctan\left(-\frac{1}{3}\right)+k\pi\end{matrix}\right.\)
//Hoặc có thể giải như sau:
\(\Leftrightarrow1-sin^2x-sinx.cosx-2sin^2x-1=0\)
\(\Leftrightarrow3sin^2x+sinx.cosx=0\)
\(\Leftrightarrow sinx\left(3sinx+cosx\right)=0\)
\(\Leftrightarrow...\)
c/
\(\Leftrightarrow1-sin^2x+\sqrt{3}sinx.cosx-1=0\)
\(\Leftrightarrow\sqrt{3}sinx.cosx-sin^2x=0\)
\(\Leftrightarrow sinx\left(\sqrt{3}cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\\sqrt{3}cosx=sinx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
d/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow2\sqrt{2}\left(tanx+1\right)=\frac{3}{cos^2x}+2\)
\(\Leftrightarrow2\sqrt{2}tanx+2\sqrt{2}=3\left(1+tan^2x\right)+2\)
\(\Leftrightarrow3tan^2x-2\sqrt{2}tanx+5-2\sqrt{2}=0\)
Pt vô nghiệm
