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Math Help - Finding Exact Solutions Trig Equations

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    Finding Exact Solutions Trig Equations

    Okay, the equation is sin 4x = -2 sin 2x. I manipulate and rearrange the equation to acquire 2 sin 2x(cos 2x + 1) = 0. I set each factor to zero, but come across a problem when setting the factor 2 sin 2x to zero. How I did, which was different from the answer key, was take the angle that makes sin 2x = 0, which was 0, and add 2npi.
    So: 2x = 0 + 2npi ---> x = npi. But they have x = (n/2)pi, giving them two more angles than I had produced--which were 0 and pi. Why did they do that? They want me to find all of the angles on the interval [0, 2pi)
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    Re: Finding Exact Solutions Trig Equations

    Can't you say \sin 2x = 2\sin x \cos x
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    Okay, the equation is sin 4x = -2 sin 2x. I manipulate and rearrange the equation to acquire 2 sin 2x(cos 2x + 1) = 0. I set each factor to zero, but come across a problem when setting the factor 2 sin 2x to zero. How I did, which was different from the answer key, was take the angle that makes sin 2x = 0, which was 0, and add 2npi.
    So: 2x = 0 + 2npi ---> x = npi. But they have x = (n/2)pi, giving them two more angles than I had produced--which were 0 and pi. Why did they do that? They want me to find all of the angles on the interval [0, 2pi)
    sin(4x)=2sin(2x)

    2sin(2x)=-2sin(2x)

    4sin(2x)=0

    sin(2x)=0
    .
    .
    .
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    Okay, the equation is sin 4x = -2 sin 2x. I manipulate and rearrange the equation to acquire 2 sin 2x(cos 2x + 1) = 0. I set each factor to zero, but come across a problem when setting the factor 2 sin 2x to zero. How I did, which was different from the answer key, was take the angle that makes sin 2x = 0, which was 0, and add 2npi.
    So: 2x = 0 + 2npi ---> x = npi. But they have x = (n/2)pi, giving them two more angles than I had produced--which were 0 and pi. Why did they do that? They want me to find all of the angles on the interval [0, 2pi)
    \displaystyle \begin{align*} \sin{4x} &= -2\sin{2x} \\ 2\sin{2x}\cos{2x} &= -2\sin{2x} \\ 2\sin{2x}\cos{2x} + 2\sin{2x} &= 0 \\ 2\sin{2x}\left(\cos{2x} + 1\right) &= 0 \\ \sin{2x} &= 0 \textrm{ or } \cos{2x} = -1 \\ 2x &= n\pi \textrm{ or } 2x = (2m + 1)\pi \textrm{ where }m, n \in \mathbf{Z} \\ x &= \frac{n\pi}{2} \textrm{ or } x = \frac{(2m + 1)\pi}{2} \\ x &= \left\{ 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \right\} \textrm{ in the interval }x \in [0, 2\pi) \end{align*}
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    Re: Finding Exact Solutions Trig Equations

    I am sorry Prove It, but I am quite illiterate regarding mathematical symbols. My problem is just that I don't understand why it is x = (n/2)pi and not x = npi
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    I am sorry Prove It, but I am quite illiterate regarding mathematical symbols. My problem is just that I don't understand why it is x = (n/2)pi and not x = npi
    Well you will need to get literate, it's only going to get more difficult as time goes on so fluency in mathematical language will help you to keep up.

    Anyway, think of the unit circle. Where is \displaystyle \begin{align*} \sin{\theta} = 0 \end{align*} ?
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    Re: Finding Exact Solutions Trig Equations

    Well it is just a simple case of never being taught them in school. Sin theta is 0 at 0, pi, and 2pi; but we can't use 2pi because the interval [0, 2pi) doesn't include it.
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    Well it is just a simple case of never being taught them in school. Sin theta is 0 at 0, pi, and 2pi; but we can't use 2pi because the interval [0, 2pi) doesn't include it.
    And if you kept going, you would have \displaystyle \begin{align*} 3\pi, 4\pi, 5\pi, \dots \end{align*} and in the other direction \displaystyle \begin{align*} -\pi, -2\pi, -3\pi, \dots \end{align*} , in other words, all the integer multiples of \displaystyle \begin{align*} \pi \end{align*} .

    We can write this in a more compact way as \displaystyle \begin{align*} n\pi \end{align*} where \displaystyle \begin{align*} n \end{align*} is an integer.
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    Re: Finding Exact Solutions Trig Equations

    Yes, that part I understand; I understand the function of the integer n. It's just--the way that I worked it out I got n times pi, but in the book they got n/2 times pi.
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    Yes, that part I understand; I understand the function of the integer n. It's just--the way that I worked it out I got n times pi, but in the book they got n/2 times pi.
    That's because you worked it out incorrectly.

    Say you had \displaystyle \begin{align*} \theta = 2x \end{align*} , then

    \displaystyle \begin{align*} \sin{\theta} &= 0 \\ \theta &= n\pi \textrm{ as we found from our discussion above} \\ 2x &= n\pi \\ x &= \frac{n\pi}{2} \end{align*}

    The reason you got the answer wrong is because you forgot about the odd multiples of \displaystyle \begin{align*} \pi \end{align*} when solving \displaystyle \begin{align*} \sin{\theta} = 0 \end{align*}
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    Re: Finding Exact Solutions Trig Equations

    I understandly perfectly what you are saying. It is just sort of contentious when comparing it to the way my professor taught me. She would tell me that if sin 2x = 0
    then 2x = 0 + 2npi ----> x = (0 + 2npi)/2 ------> x =npi. Then, I would continue to input values in for n, until I had all of the angles in the interval [0, 2pi) That's why I don't understand why it would be (n/2)pi.
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    Re: Finding Exact Solutions Trig Equations

    Quote Originally Posted by Bashyboy View Post
    I understandly perfectly what you are saying. It is just sort of contentious when comparing it to the way my professor taught me. She would tell me that if sin 2x = 0
    then 2x = 0 + 2npi ----> x = (0 + 2npi)/2 ------> x =npi. Then, I would continue to input values in for n, until I had all of the angles in the interval [0, 2pi) That's why I don't understand why it would be (n/2)pi.
    Then you have misunderstood what your professor has said. I have explained why it is \displaystyle \begin{align*} \frac{n\pi}{2} \end{align*} , arguing about it won't make what you think you heard correct.

    What she would have said is that when you solve a trigonometric equation, you need to find the solution from the TWO QUADRANTS that satisfy the equation, then add multiples of \displaystyle \begin{align*} 2\pi \end{align*} .

    IF this is the case, then in this special case, where \displaystyle \begin{align*} \sin{\theta} = 0 \end{align*} , you would find that \displaystyle \begin{align*} \theta = \{0, \pi\} + 2n\pi \end{align*} , and when you write down the first few values, you'll see that the pattern gives \displaystyle \begin{align*} n\pi \end{align*} . You are incorrect because each time you have forgotten about the solution \displaystyle \begin{align*} \theta = \pi \end{align*} and all of the odd integer multiples.
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