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tammyhawatmeh

  • 3 years ago

cos θ = − root2 /2

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  1. baldymcgee6
    • 3 years ago
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    how can you get rid of the cosine?

  2. tammyhawatmeh
    • 3 years ago
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    arc cos?

  3. baldymcgee6
    • 3 years ago
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    yep

  4. baldymcgee6
    • 3 years ago
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    do you know the unit circle?

  5. tammyhawatmeh
    • 3 years ago
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    yup

  6. baldymcgee6
    • 3 years ago
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    okay, so where is cos(θ) = -sqrt(2/2) ?

  7. tammyhawatmeh
    • 3 years ago
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    3pi/4

  8. baldymcgee6
    • 3 years ago
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    Yes.. correct.

  9. hartnn
    • 3 years ago
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    what about 5pi/4 ?

  10. hartnn
    • 3 years ago
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    now do u need general solution or within the range 0 to 2pi ?

  11. tammyhawatmeh
    • 3 years ago
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    these are the directionsSolve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)

  12. hartnn
    • 3 years ago
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    means genearalized solution, can u generalize this ?

  13. tammyhawatmeh
    • 3 years ago
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    5pi/4 ,-5pi/4,3pi/4,-3pi/4

  14. baldymcgee6
    • 3 years ago
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    There is an infinite amount of solutions, you need to write it in a way to GENERALIZE the solution.

  15. hartnn
    • 3 years ago
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    so let me generalize it. cos function is periodic with 2pi. so values repeat after every 2pi*k here, \(\theta =\pm(3\pi/4+2\pi k),\pm(5\pi/4+2\pi k)\)

  16. tammyhawatmeh
    • 3 years ago
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    ohh i see but why do u add the 2pik at the end ? because it is a periodic function?

  17. hartnn
    • 3 years ago
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    yes, as i said cos function is periodic with 2pi. so if a is solution, so is a+2pi, a-2pi,a+4pi,a-4pi, and so on

  18. tammyhawatmeh
    • 3 years ago
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    ohh i see i missed this whole lesson on this section and i dont know any of it

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