<s>[INST] determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$ Determine the roots of the equation $7\sec^2{x}+2\tan{x}-6=2\sec^2{x}+2$, when $0≤x≤2π$. I don't know how to solve this equation algebraically. Please help?
 [/INST] Using $1+tan^2=sec^2$,
$$7\sec^2(x)+2\tan(x)-6=2\sec^2(x)+2\iff $$
$$7+7\tan^2(x)+2\tan(x)-6=2+2\tan^2(x)+2\iff $$
$$5\tan^2(x)+2\tan(x)-3=0 \iff$$
$$(\tan(x)+1)(5\tan(x)-3)=0 \iff$$
$$\tan(x)=-1 \hbox{  or  } \tan(x)=3/5.$$
Thus, in $[0,2\pi)$, $x=3\pi/4$ or $x=7\pi/4$ or $x=\tan^{-1}(3/5)$ or $x=\tan^{-1}(3/5)+\pi.$
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