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anonymous

  • one year ago

Find all solutions in the interval [0, 2pi). tan x + sec x = 1

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  1. anonymous
    • one year ago
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    MY WORK: (sin x/ cos x) + (1/ cos x) = 1 (sin x + 1)/ cos x = 1 Then, I'm stuck. :(

  2. Vocaloid
    • one year ago
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    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?

  3. Vocaloid
    • one year ago
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    ah, a bit of a typo last expression should be sec^2(x) - 1 = 1 - 2secx + sec^2(x)

  4. anonymous
    • one year ago
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    So what I was doing is wrong?

  5. Vocaloid
    • one year ago
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    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)

  6. Vocaloid
    • one year ago
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    there might be a way to do it your way, but I can't think of an easy way to go from there

  7. anonymous
    • one year ago
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    Wait, so will it be equal to 0?

  8. Vocaloid
    • one year ago
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    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

  9. anonymous
    • one year ago
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    Then it is 0. :D

  10. Vocaloid
    • one year ago
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    yea

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