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cassiekola
find all solutions of the equation in the terminal [0, 2 pi). sec^2 x + tan x = 1
Use the fact that \[sec^{2} x = \tan^{2}x + 1\] and therefore the equation becomes, after substitution and some algebraic manipulation, \[\tan^{2}x +\tan x = 0\] Then factor to get \[\tan x(\tan x +1)=0\] and therefore either \[tan x = 0\] or \[ \tan x + 1=0\] In the first case, then on this interval, either \[x = 0\] or \[x = \pi\] In the second case, then we get \[ \tan x = -1\] and therefore \[x=\frac{3\pi}{4}\] or \[x=\frac{7\pi}{4}\].
Well: \[\sec^2(x)=1+\tan^2(x) \implies \tan^2(x)+\tan(x)+1=1\] \[ \implies \tan(x)(\tan(x)+1)=0 \implies \tan(x)=0, \tan(x)=-1; x=n \pi, x=\frac{(4n+3) \pi}{4}\]
The last part should be: \[\frac{(4n+3) \pi}{4}\]