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Find all solutions to Sin 2x = 2Tan 2x

  1. Jun 28, 2005 #1
    Is this correct??

    Hello All

    I need to find all solutions to the following did i do it correct?

    Sin 2x = 2Tan 2x

    2Sinx Cosx = 2(2Tanx / 1-Tan^2x)

    Sinx Cosx = (2(Sinx / Cosx)/(cos^2x-Sin^2x / Cos^2x))

    Sinx Cosx = 2(Sinx / Cosx) X (Cos^2x / cos^2x-Sin^2x)

    Cosx = 2 (Cosx / Cos^2x - Sin^2x)

    1 = (2 / Cos^2x - Sin^2x)

    Cos^2x - Sin^2x = 2

    1 - Sinx^2 - Sin^2x = 2

    2Sin^2x = -1
    Sin^2x = -1/2

    Sinx = -1/SQROOT(2)



    Is this correct??
     
  2. jcsd
  3. Jun 28, 2005 #2

    arildno

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    Nope.
    Remember that Tan(2x)=Sin(2x)/Cos(2x)

    Thus, you may rewrite your original equation as:
    [tex]\sin(2x)(1-\frac{2}{\cos(2x)})=0[/tex]

    EDIT:
    Find out the requirements for either factor to be zero; you should find that one of those requirements is impossible to achieve.
     
    Last edited: Jun 28, 2005
  4. Jun 28, 2005 #3

    SGT

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    You certainly made a mistake. You arrived to the equation:
    [tex]cos^2x - sin^2x = 2[/tex]
    but
    [tex]cos^2x - sin^2x = cos 2x[/tex]
    So you have
    cos 2x = 2
    This has no real solution.
    proceeding your development you got
    [tex]sin^2x = -\frac{1}{2}[/tex]
    whose solution is
    [tex]sin x = -\frac{i}{\sqrt{2}}[/tex]
     
  5. Jun 28, 2005 #4
    Let's call tan(x) "t"

    then we have 2*t/(1+t^2)=4*t/(1-t^2)

    we must exclude that t=+-1 to give the expression a meaning. (x!=pi/4+kpi/2)

    then we obtain one solution: t=0 (x=kpi)
    we have if t!=0: 1-t^2=2+2t^2 that is 3t^2=-1, which has no real solution (anyway you could be interested in finding the complex ones)
    So the only real solution is x=k*pi, where k belongs to Z
     
  6. Jun 28, 2005 #5
    Arildno Which side are you talking about?

    SGT: Did have COS2x = 2 but was told by a classmate that I did the problem wrong. I kind of get confussed with the 2x in the statment. What does it really mean.
     
  7. Jun 28, 2005 #6

    SGT

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    I did not follow all your development because your notation is almost unintelligible, but I think your classmate is right. You probably made some mistake.
    The 2x means the double of the arc x. If [tex]x = \frac{\pi}{3}[/tex] for instance, then [tex]2x = \frac{2\pi}{3}[/tex]
     
  8. Jun 29, 2005 #7

    arildno

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    1. Our problem is to determine the set of x-values that makes
    [tex]\sin(2x)=2\tan(2x)[/tex]
    into a TRUE statement.
    That is, we want to find the "solutions" to that equation.

    2. We have, for all x-values the identity [tex]\tan(2x)=\frac{\sin(2x)}{\cos(2x)}[/tex]
    where with "all x-values" should be understood all real values except those for which [tex]\cos(2x)=0[/tex] (i.e, when tan(2x) is infinite).

    3. Given the identity in 2., we may rephrase our original problem into finding the solution set for the following equation:
    [tex]\sin(2x)=2\frac{\sin(2x)}{\cos(2x)}[/tex]

    4. Now, adding any arbitrary number to both sides of an equation won't change the solution set we're after, so by adding -sin(2x) to both sides, we can rephrase our problem into determining the solution set to the following equation:
    [tex]2\frac{\sin(2x)}{\cos(2x)}-\sin(2x)=0[/tex]

    5) Arbitrary real numbers a,b,c fulfill the distributive law: a*(b+c)=a*b+a*c
    Recognizing the common factor sin(2x) on the left-hand side expression in the equation given under 4., we may rephrase our problem into determining the solution set of the following equation:
    [tex]\sin(2x)*(\frac{2}{\cos(2x)}-1)=0[/tex]

    6) Now, if the product of two numbers is to be zero, then at least one of the numbers themselves must be zero.
    Therefore, if the equation under 5. is to be a true statement, then we must have EITHER:
    [tex]\sin(2x)=0[/tex]
    OR:
    [tex]\frac{2}{\cos(2x)}-1=0[/tex]

    7. Let's look at the last one.
    If we are to have [tex]\frac{2}{\cos(2x)}-1=0[/tex], then this is equivalent to demanding [tex]\cos(2x)=2[/tex]
    But, since the cosine is always less than or equal to 2, we see that the statement:
    [tex]\cos(2x)=2[/tex] cannot have any solutions at all
    (that is, whatever number "x" you substitute into the expression on the left-hand side, will never make cos(2x) equal to 2.

    8.
    Thus, we may conclude, that the solution set to our original problem will be the solution set of the equation:
    [tex]\sin(2x)=0[/tex]
     
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