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Homework Help: Trig equations losing solutions

  1. Oct 28, 2005 #1
    one of the questions is as follows:

    7cosx +6sinx =2...the answer is required between 0 and 360 degrees.
    I first tried to tackle it using rsin(x+y) giving rsiny = 7, rcosy = 6...
    r = sqrt(85), y =49.4deg

    sqrt85sin(x + 49.4deg) = 2...sin(x + 49.4deg) =2/sqrt85...
    x + 49.4deg =+/- 12.53deg
    this gives rise to 323.deg and 298.1deg...of which only 323.1deg is correct, the other answer is 118.1deg but I got this by checking the answers, scratching my head for a while and then using rcos(x-y) with just the positive square root of 85.

    the book doesn't state which expression should be used and a different book implies that all of the 4 expressions will yield the same result in all situations...my question is how should I know which expression to use or is there something wrong with my working? (I do know with hindsight that the above question requires the cosine of x to be positive if the sine is negative and vice-versa)

    (I also have noticed that the wrong answer I got is a rotation about 180deg of the correct answer...but I am sure that this is not the way I should be arriving at a solution)
    Last edited: Oct 28, 2005
  2. jcsd
  3. Oct 28, 2005 #2
    i'm thinking maybe you could use trig identities? i think i may have seen one like it in my textbook but i was never very good with trig!!
  4. Oct 28, 2005 #3
    meh!...didn't subtract 12.53deg from 180deg :biggrin: hit me like a hammer driving back home...and could only get back to the computer now
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