Question

f(x)=2secx-tanx, -π/2<x<π/2

find the points at which the tangent is horizontal

Answer 1

f'=0=2secxtanx-sec^2x

secx(2tanx-sex)=0
secx=0 or tanx=1/2 secx which leads to
sinx=.5

now, because secx cannot be zero, then
sinx=.5
or x=30 deg or PI/6