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anonymous

  • 5 years ago

Solve this equation on the interval 0 θ < 2π. If there is only one solution, write it in the first box and write NONE in the second box. If there is no solution, write NONE in both boxes. Round your answer(s) to two decimal places using radians. -2 cos(θ) + 1 = 2 θ = (smaller value of θ) θ = (larger value of θ)

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  1. anonymous
    • 5 years ago
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    the interval is 0( which is less than or equal to) theta ( which is less than) 2pi

  2. anonymous
    • 5 years ago
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    -2cos(x) = 1 cos(x) = -1/2 the related angle is cos inverse of 1/2 , which is 60 degrees

  3. anonymous
    • 5 years ago
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    cos is negative in the 2 and 3 quadrants

  4. anonymous
    • 5 years ago
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    x= 180 -60 and x= 180 + 60 x= 120 , 240 degrees

  5. anonymous
    • 5 years ago
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    \[\frac{2\pi}{3} , \frac{4\pi}{3} \] in radians

  6. anonymous
    • 5 years ago
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    thanks. but the answer seems to be wrong? not sure why

  7. anonymous
    • 5 years ago
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    I am fairly sure what I did was correct

  8. anonymous
    • 5 years ago
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    could it be because the question says round the answer to two decimal points in radians?

  9. anonymous
    • 5 years ago
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    yes lol :|

  10. anonymous
    • 5 years ago
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    obviously if you entered the exact values then they wouldnt like it

  11. anonymous
    • 5 years ago
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    I assumed you would know how to convert the exact values into radians, lol, just calculator work

  12. anonymous
    • 5 years ago
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    yes, however i dont have a calc lol! i think the first one is 2(3.14)/3 ?

  13. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=cos%28x%29+%3D-1%2F2+%2C+0%3Cx%3C360 theres a calculator even solves eqns and integrals and everything

  14. anonymous
    • 5 years ago
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    I use it to check my answers

  15. anonymous
    • 5 years ago
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    ok thanks. but to be 100% sure(kind of forgot) radians is using pi as 3.14?

  16. anonymous
    • 5 years ago
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    yes approx that ,

  17. anonymous
    • 5 years ago
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    thanks mate. got it right!

  18. anonymous
    • 5 years ago
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    would you also be able to show me how to do the question just above this question?

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