TH1 cosx=0 \(\Leftrightarrow1=0\left(vl\right)\)
TH2 \(cosx\ne0\) chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow-2tan^3x-6+1+tân^2x+3tanx\left(1+tan^2x\right)=0\)
\(\Leftrightarrow tan^3x+tan^2x+3tanx-5=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+2tanx+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+2tanx+5=0\left(VN\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)
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