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jst take dy/dx

y'=sec(x)^2+sin(x), y'=0 when sec(x)^2=-sin(x), solve, profit!

thanks!!

wait how dou get sin(x)??

Oh, sin(x) was my mistake.

Apparently, I'm still bad at calculus. XP It should be (sin(x))(cos(x)^-2).

i thought derivative of sec x is sec x tan x

yea........

sin(x)/cos(x)^2=tan(x)/cos(x)=tan(x)sec(x)

Same thing. I just didn't remember it, I had to prove it out.

ok so what i do now after i got the derivative ??

sec(x)^2+tan(x)sec(x)=y'=0, tan(x)+1=0, tan(x)=-1, x=arctan(-1)

Woops, my bad. Wait a sec.

(sec(x)+tan(x))=0, tan(x)=-sec(x)

there we go. when do tan(x) and -sec(x) intersect?

wait a minute i need to look at unit circle, i dont have those memorized

sin(x)/cos(x)=-1/cos(x), sin(x)=-1 when is this true?

i think 3pi/4, 7pi/4