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B Finding the smallest positive solution to trig equation

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  1. Aug 12, 2016 #1
    I have the equation ##\sin 3x = \cos 7x##, and, in degrees, I have to find the smallest positive solution.

    Immediately, we can see that sin and cos are equal if their arguments are complements, so ##3x + 7x = 90##, which means that ##x = 9##.

    I know that that is a correct solution, but how do I show that it is, in fact, the smallest positive solution?
     
  2. jcsd
  3. Aug 12, 2016 #2
    I would do it this way:
    sin(x)=cos(x) if x=45 degrees. Of course, you knew that and could then get the correct solution. However, you are missing one part -- x=45+180n degrees where n is an integer. If you are looking for the smallest possible x, n would have to be zero.
     
  4. Aug 12, 2016 #3

    Charles Link

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    An interesting problem. Since the angles are both in the first quadrant, I think you clearly have the smallest x. To prove it might take a little work, but it would take more effort to find the second smallest or 3rd smallest. (Solving the equivalent ## \sin(3x)=\sin(90-7x) ##, trigonometric identities allow 360 degrees to be input into either side and you have essentially the same equation ) e.g. If ## 3x=360+(90-7x) ## so that ## 10x=450 ## then x=## 45 ##. I found another solution, but is it the second smallest? editing... You can also add 720 or other multiples of 360 to either side, or you can do ## 180-\theta ## to either term inside the sine function without changing the equation...
     
    Last edited: Aug 12, 2016
  5. Aug 13, 2016 #4

    Charles Link

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    A follow-on to post #3: Overlaying a graph of y =sin(3x) and a graph of y=cos(7x) is perhaps the quickest way to see the (approximate) solutions of this problem. It will show you that your x=9 is the smallest x.
     
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