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  • bonkysleuth
    bonkysleuth's Avatar
    314 posts since Mar '07
    • 27 Aug `08, 6:14PM

      Encountered a little problems with this question. Please take a look.(=

      find all the angles between 0 deg and 360 deg which satisfy the equation

       

      2sin y = tan y

      cos y = 2

      y lies in first or fourth quadrant.

      cos alfa = 1/2

      alfa = 60 deg

      y - 60 deg, 300 deg

       

      I am missing on 180 deg(stated in the answers). how can you find that?

       

      THANKS!

  • Moderator
    eagle
    eagle's Avatar
    17,973 posts since Aug '01
    • 27 Aug `08, 11:53PM

      2sin y = tan y
      2sin y - sin y / cos y = 0

      sin y (2cos y - 1) = 0

      either siny = 0 or cos y = 1/2

       

      From sin y = 0, you get your 180 degrees.

       

      Remember for such equations, never divide the sin y away. You will lose solutions, as shown clearly by your example.

      Another example would be  2x^2 - x = 0
      You don't divide by x to get 2x -1 = 0. Instead, you factorise to x (2x - 1) = 0. The same thing is to be used on your question here.

       

      Regards,
      Eagle
      Owner of Strategic Tuition

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