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Need help with trigonometric equation

  1. Mar 18, 2009 #1
    Hi All,
    I’m currently attempting to work ahead on my precalculs and I’m looking at trigonometric equations.

    I seem to have a bit of a problem with this example (i haven’t had too many issues with the rest of the exercises),
    9cos(2x)+sin(x)=9 (solve for x in the interval 0 <=x <=2pi.)
    I’m thinking that I could move 9 on the other side and hence the equation will equal to 1:

    I know that cos(2x)=cos2x-sin2x and than the equation should look like this:
    Am i on the right track? What is the next step form here?
    Thank you in advance.
  2. jcsd
  3. Mar 18, 2009 #2


    User Avatar
    Gold Member

    Woa be careful what you do there; If you are going to divide by 9, make sure you divide EVERYTHING by 9, so it becomes:

    cos(2x) + (1/9)sin(x)=1

    Anyway, other than that, i would probably proceed in the same way as you did. With these types of questions, it is often easiest to reduce the equation to a form such that only one type of trig function is present. Can you see how a common identity can be used to further reduce the equation down to one with only sine's? What common type of equation does this then resemble?
  4. Mar 18, 2009 #3
    Thank you for replaying danago.

    That definitely helps. So after dividing everything by 9 I’m getting:
    I can use the double angle formula to reduce the equation to one type of trig function, hence:
    And i now have a quadratic equation type 2x^2+(1/9)x
    I can use the quadratic formula and I’m nearly done.

    Many thanks.
  5. Mar 18, 2009 #4


    Staff: Mentor

    Two comments:
    When you write cos(2x)+1/9sin(x)=1, some people might (incorrectly) take the sine term to be 1/(9sin(x)). You can write this more clearly as 1/9 * sin(x).

    You can use the quadratic formula to solve -sin^2(x)+1/9sin(x)=0, but it's quicker and simpler just to factor sin(x) from each term to get sin(x)(-sin(x) + 1/9) = 0, and then set each factor to 0 to solve for sin(x) and then x.
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