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anonymous

  • one year ago

Is this correct? Solve the equation on the interval of \( 0\geq \theta < 2\pi \) \( 2cos \theta + 3 = 2 \) \( 2 cos \theta = 2-3 \) \( 2 cos \theta = -1 \) \( \frac{2 cos \theta}{2} = \frac{-1}{2} \) \( cos \theta = \frac{-1}{2} \) This was on my midterm and I made an A on it but this one I got wrong but this looks correct? What did I do wrong?

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  1. anonymous
    • one year ago
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    Solve the equation on the interval of \( 0\geq \theta < 2\pi \) \[ \huge 2\cos \theta + 3 = 2 \] \[ \huge 2 \cos \theta = 2-3 \] \[\huge 2 \cos \theta = -1 \] \[ \huge \frac{2 \cos \theta}{2} = \frac{-1}{2} \] \[\huge \cos \theta = -\frac{1}{2} \]

  2. Curry
    • one year ago
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    wait how many answers did you provide? there are two answers.

  3. anonymous
    • one year ago
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    One sec it should be \(\huge 0\leq \theta < 2\pi \)

  4. anonymous
    • one year ago
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    I provided one

  5. anonymous
    • one year ago
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    but if we work out the equation, like I did, it comes to what I have

  6. TheSmartOne
    • one year ago
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    You need to solve for \(\sf\theta\)

  7. anonymous
    • one year ago
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    Ok so he wanted the angle

  8. TheSmartOne
    • one year ago
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    Yes, and you should get 2 answers for the angle

  9. anonymous
    • one year ago
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    Yes I get two. I was thinking he just wanted us to do it like that because some of our homework does it like I just did it.

  10. TheSmartOne
    • one year ago
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    What 2 angles did you get? :)

  11. DanJS
    • one year ago
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    cos(a) = x on the unit circle with radius 1... picture that thing

  12. anonymous
    • one year ago
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    I got 120 and 240 degrees. or 2pi/3 and 4pi/3

  13. TheSmartOne
    • one year ago
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    |dw:1435870657418:dw|

  14. TheSmartOne
    • one year ago
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    Correct :)

  15. anonymous
    • one year ago
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    Yep that is it :-) Just a silly mistake on my part :-( oh well lol

  16. anonymous
    • one year ago
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    Got the hardest one right because I put the angle lol

  17. TheSmartOne
    • one year ago
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    If they didn't have the restrictions on the value of \(\sf\theta\) then the answer would be: \(\sf\Large \theta=\frac{2\pi}{3}\pm2\pi n\) \(\sf\Large \theta=\frac{4\pi}{3}\pm2\pi n\)

  18. anonymous
    • one year ago
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    Funny, I made the same mistake on two but for some reason, on the hardest one I put the angle for that one but not the other two.... . DUFF what was I thinking lol

  19. anonymous
    • one year ago
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    Anyways... I still have an A in the class and I made an A on the midterm so I am happy :-)

  20. TheSmartOne
    • one year ago
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