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Math Help - How to solve -2sin(t)+ cos(t) =0

  1. #1
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    How to solve -2sin(t)+ cos(t) =0

    I have: -2sin(t)+ cos(t) =0

    How can I find out what t is?

    I dont even know how to start, because I havent done this in over 3 years..

    Help would be appreciated!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tinyone View Post
    I have: -2sin(t)+ cos(t) =0

    How can I find out what t is?

    I dont even know how to start, because I havent done this in over 3 years..

    Help would be appreciated!
    first note that neither sin(t) nor cos(t) can be zero here (do you see why?)

    since they are not zero, we can divide by either of them. hence,

    \displaystyle -2 \sin t + \cos t = 0

    \displaystyle \Rightarrow 2 \sin t = \cos t

    \displaystyle \Rightarrow 2 \tan t = 1 ..............divided by the cos(t)

    \displaystyle \Rightarrow \tan t = \frac 12

    Now what?
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  3. #3
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    Just a word of note: t may have more than one value.
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  4. #4
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    t=arctan(1/2) = 0.463647609

    But i need the answer to be in π (pi)...

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  5. #5
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    This isn't a special angle so you won't get it in terms of pi
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  6. #6
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    wonderboy 1953, how can I find the other solution...

    is it perhaps done this way: 2π - 0.463647609

    ?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tinyone View Post
    wonderboy 1953, how can I find the other solution...

    is it perhaps done this way: 2π - 0.463647609

    ?
    don't approximate arctan(1/2), leave it as arctan(1/2)

    there are infinitely many solutions, not just 2--not unless the problem itself tells you to restrict your answer.

    Hint 1: tangent is periodic.

    Hint 2: tangent takes on positive values in the first and third quadrants. arctan(1/2) is in the first quadrant. apply what you know of reference angles.
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