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anonymous

  • 5 years ago

i need to solve this over the interval [0,2pi]. tan^2x=2secx-1

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  1. anonymous
    • 5 years ago
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    tan^2 x - 2 sec x +1

  2. anonymous
    • 5 years ago
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    i simplified tan^2x into sec^2x-1. is that a good step?

  3. anonymous
    • 5 years ago
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    (sec^2 x -1) - 2 sec x +1 sec^2 x - 2 sec x

  4. anonymous
    • 5 years ago
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    sec (x)=u so substituting u^2-2u=0 We may complete the square by adding 1 to both side u^2 -2u+1=1 (u-1)(u-1)=1 u=2 sec (x)=2 so since 1/sec(x)= cos(x) cos(x)=1/2 x=pi/3

  5. anonymous
    • 5 years ago
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    wait what is u?

  6. anonymous
    • 5 years ago
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    I just define u for sec(x) for simplicity

  7. anonymous
    • 5 years ago
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    ok so is the work with sec(x)=u a continuation of the post you did before that?

  8. anonymous
    • 5 years ago
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    U is used in place of sec(x), just to make it easier to see

  9. anonymous
    • 5 years ago
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    no, I did not need to use U. I could have instead solve it using sec(x)

  10. anonymous
    • 5 years ago
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    Solved without using u sec(x)^2-2sec(x)=0 We may complete the square by adding 1 to both side sec(x)^2 -2sec(x)+1=1 (sec(x)-1)(sec(x)-1)=1 sec(x)=2 sec (x)=2 so since 1/sec(x)= cos(x) cos(x)=1/2 x=pi/3

  11. anonymous
    • 5 years ago
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    could i have used the quadratic formula to solve from where it was secx^2-2secx=0?

  12. anonymous
    • 5 years ago
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    and also how do you know you have to complete the square?

  13. anonymous
    • 5 years ago
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    Yes, but it is much easier

  14. anonymous
    • 5 years ago
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    u^2-2u=0 I could see that by adding 1 to left side, I would be able to factor easily. But if you add something one one side, same must be done on the other side

  15. anonymous
    • 5 years ago
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    thank you!

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