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anonymous

  • 5 years ago

sin²θ-cos²θ-cosθ=1. Find all solutions in the interval [0,2π).

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  1. anonymous
    • 5 years ago
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    \[sin^2(x)-cos^2(x)-cos(x)=1\] first rewrite in term of cosine alone, using the identity \[sin^2(x)+cos^2(x)=1\] \[1-cos^2(x)-cos^2(x)-cos(x)=1\] \[1-2cos^2(x)-cos(x)=1\] \[-2cos^2(x)-cos(x)=0\] \[2cos^(x)+cos(x)=0\] \[cos(x)(2cos(x)+1)=\] \[cos(x) = 0\] or \[2cos(x)+1=0\]

  2. anonymous
    • 5 years ago
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    of course you still have to solve \[cos(x) = 0\] \[x=\frac{\pi}{2}\]or \[x=\frac{3\pi}{2}\]

  3. anonymous
    • 5 years ago
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    \[2cos(x)+1=0\] \[cos(x)=-\frac{1}{2}\] \[x=\frac{2\pi}{3}\] \[x=\frac{4\pi}{3}\]

  4. anonymous
    • 5 years ago
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    so, x=2π/3 and 4π/3?

  5. anonymous
    • 5 years ago
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    you have 4 solutions all together.

  6. anonymous
    • 5 years ago
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    oh, ok...also the π/2 and 3π/2.... thanks

  7. anonymous
    • 5 years ago
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    right. is it clear?

  8. anonymous
    • 5 years ago
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    it is, when someone shows you!

  9. anonymous
    • 5 years ago
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    mmm rly u have unfintly solutin here by adding the period of ur functin

  10. anonymous
    • 5 years ago
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    of course! idea is to write using one trig function, then you get a quadratic equation to solve. there are an infinite number of solutions, that is why the question restricts interval from \[[0,2\pi)\]

  11. anonymous
    • 5 years ago
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    i.e ur sol + 2n bay

  12. anonymous
    • 5 years ago
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    thanks 'satellite 73' . hard to know where to start.

  13. anonymous
    • 5 years ago
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    welcome!

  14. anonymous
    • 5 years ago
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    mm sry 'm jst dont care 2 the question

  15. anonymous
    • 5 years ago
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    it is, when someone shows you!

  16. anonymous
    • 5 years ago
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    oh, ok...also the π/2 and 3π/2.... thanks

  17. anonymous
    • 5 years ago
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    so, x=2π/3 and 4π/3?

  18. anonymous
    • 5 years ago
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    yes 4 total. two for the equation \[cos(x)=0\] and two for \[cos(x)=-\frac{1}{2}\]

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