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moongazer

  • 4 years ago

I. Find the solutions of the following trigonometric conditional equations for 0 ≤ ϴ ≤ 2π. (Rx2) 1. 4 cos ϴ + 4 = 0 2. sec 2 ϴ = 1 3. sin ϴ - 1 = 0 4. 2 cos ϴ - 1 = 0 5. 4 cos2 ϴ = 3 6. cos ϴ + sin 2ϴ = 0 7. cot 2ϴ = 3 8. cos2 ϴ - cos ϴ = 0 9. sin2 ϴ = ¾ 10. csc2 ϴ - 1 = 0

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  1. moongazer
    • 4 years ago
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    I don't know how to solve the parts with sec 2 ϴ = 1,sec 2 ϴ = 1, csc2 ϴ - 1 = 0 I don't know if it is a typo or it is the real equation.

  2. dumbcow
    • 4 years ago
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    is it sec^2 or sec(2x) ?

  3. moongazer
    • 4 years ago
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    What do you think?

  4. moongazer
    • 4 years ago
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    is it possible to solve if it is sec(2x) ????

  5. moongazer
    • 4 years ago
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    because I think it is typo

  6. dumbcow
    • 4 years ago
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    yes sec(2x) = 1 1/cos(2x) =1 cos(2x) = 1 cos is 1 when angle is 0 2x = 0 x = 0

  7. moongazer
    • 4 years ago
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    thanks! I'll just try both

  8. dumbcow
    • 4 years ago
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    if its squared, then it will be sec(x) = +-1

  9. moongazer
    • 4 years ago
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    w8 I think your solution is incomplete

  10. dumbcow
    • 4 years ago
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    oh, you are right 2x =2pi x = pi

  11. dumbcow
    • 4 years ago
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    cos^2 + sin^2 = 1

  12. moongazer
    • 4 years ago
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    isn't it that sec 2A will be 1/(cos^2 A - sin^2 A) ??????????

  13. moongazer
    • 4 years ago
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    My teacher taught us that cos2A = cos^2 A - sin^2 A

  14. dumbcow
    • 4 years ago
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    sorry, cos^2 - sin^2 = 1 cos^2 = 1+ sin^2 cos^2 = 1+(1-cos^2) 2cos^2 = 2 cos^2 = 1 cos A = +-1 so you end up with same solutions A = 0,pi

  15. moongazer
    • 4 years ago
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    ohhh, thanks!

  16. dumbcow
    • 4 years ago
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    sure, the csc one will work the same way

  17. moongazer
    • 4 years ago
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    another thing is 1/x = 1 also the same as x=1 ??? because when you cross multiply it, it will be the result?

  18. dumbcow
    • 4 years ago
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    correct

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