anonymous
  • anonymous
Find all solutions in the interval [0, 2pi). tan x + sec x = 1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
MY WORK: (sin x/ cos x) + (1/ cos x) = 1 (sin x + 1)/ cos x = 1 Then, I'm stuck. :(
Vocaloid
  • Vocaloid
subtract sec(x) from both sides first tan(x) = 1 - sec(x) then square both sides tan^2(x) = (1-secx)^2 = 1 - 2secx + sec^x then use the trigonometric identity tan^(x) = sec^2(x)-1 to change the left side sec^2(x) - 1 = 1 - 2secx + sec^x can you finish from here?
Vocaloid
  • Vocaloid
ah, a bit of a typo last expression should be sec^2(x) - 1 = 1 - 2secx + sec^2(x)

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anonymous
  • anonymous
So what I was doing is wrong?
Vocaloid
  • Vocaloid
well, not "wrong" but it would be difficult to get an answer by converting tan(x) and sec(x) into sin(x) and cos(x)
Vocaloid
  • Vocaloid
there might be a way to do it your way, but I can't think of an easy way to go from there
anonymous
  • anonymous
Wait, so will it be equal to 0?
Vocaloid
  • Vocaloid
sec^2(x) - 1 = 1 - 2secx + sec^2(x) cancel out sec^2(x) -1 = 1 - 2sec(x) 2sec(x) = 2 sec(x) = 1 then solve for x
anonymous
  • anonymous
Then it is 0. :D
Vocaloid
  • Vocaloid
yea

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