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Math Help - Trig equation, is this correct?

  1. #1
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    Trig equation, is this correct?

    Solve the following equations in the given intervals.

     sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1

     0\leq \theta \leq 2\pi
     sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1

     1 + tan^2 \theta = sec^2 \theta

     1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1

     tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0

     tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3}

     tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}

    so

     tan \theta = -\sqrt{3}

    or

     tan \theta = 1

    is this method correct? thanks!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tweety View Post
     sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1

     1 + tan^2 \theta = sec^2 \theta

     1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1

     tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0

     tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3}

     tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}

    so

     tan \theta = -\sqrt{3}

    or

     tan \theta = 1

    is this method correct? thanks!
    it is wrong
    let tan(\theta)=t and solve it by t=\frac{-b\mp \sqrt{b^2-4ac}}{2a} in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

    then

    tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}

    \theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)
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  3. #3
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    Quote Originally Posted by Amer View Post
    it is wrong
    let tan(\theta)=t and solve it by t=\frac{-b\mp \sqrt{b^2-4ac}}{2a} in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

    then

    tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}

    \theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)
    Hi,

    I dont understand how you got to the quadratic equation?

    the correct answers are,  \frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3}
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  4. #4
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    tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}<br />

    You can't factorise when it's like that, you can only factorise when one side is equal to 0

    <br />
tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0<br />

    You can use the quadratic equation from there. If it helps let u = tan \theta
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tweety View Post
    Hi,

    I dont understand how you got to the quadratic equation?

    the correct answers are,  \frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3}
    tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0

    let t=tan(theta)

    t^2 -(1+\sqrt{3})t+\sqrt{3}=0

    t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}

    t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}

    t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}
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  6. #6
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    Quote Originally Posted by Amer View Post
    tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0

    let t=tan(theta)

    t^2 -(1+\sqrt{3})t+\sqrt{3}=0

    t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}

    t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}

    t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}
    Oh right, I get it now thanks
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  7. #7
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     \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0

    (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0

    \tan{\theta} = \sqrt{3} ... \theta = \frac{\pi}{3} , \frac{4\pi}{3}<br />

    \tan{\theta} = 1 ... \theta = \frac{\pi}{4} , \frac{5\pi}{4}
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  8. #8
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    Quote Originally Posted by skeeter View Post
     \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0

    (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0

    \tan{\theta} = \sqrt{3} ... \theta = \frac{\pi}{3} , \frac{4\pi}{3}<br />

    \tan{\theta} = 1 ... \theta = \frac{\pi}{4} , \frac{5\pi}{4}
    how did you get this expression  (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0
    from
     \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0

    ? is it a difference of two square??
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  9. #9
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    Quote Originally Posted by Tweety View Post
    how did you get this expression  (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0
    from
     \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0

    ? is it a difference of two square??
    Suppose you had x^2- (1+ a)x+ a. Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?
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  10. #10
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Tweety View Post
    how did you get this expression  (\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0
    from
     \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0

    ? is it a difference of two square??
    look at this

    x^2-5x+6 \Rightarrow (x-2)(x-3) -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

    x^2-4x+3\Rightarrow (x-1)(x-3) -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

    like this or use

    x=\frac{-b\mp \sqrt{b^2-4ac}}{2a} if you do not understand it
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  11. #11
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    Difference of 2 squares here is not needed.

    Skeeter has used the fact that -\sqrt{3}\times -1 = \sqrt{3}

    and that -\sqrt{3}+ -1 = -(\sqrt{3}+1) like what you would do when solving an ordinary quadratic that has 1 as the coeffecient for the x^2 term
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  12. #12
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    Quote Originally Posted by HallsofIvy View Post
    Suppose you had x^2- (1+ a)x+ a. Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?

    if you had the expression  (x-1)(x-a) this expands out to  x^2-ax-x +a

    right?

    Than you can't really factor it any further to get back to the original expresssion. which is why I can't understand how it turn outs to be (x-1)(x-a)
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  13. #13
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    Quote Originally Posted by Amer View Post
    look at this
    Quote Originally Posted by Amer View Post

    x^2-5x+6 \Rightarrow (x-2)(x-3) -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

    x^2-4x+3\Rightarrow (x-1)(x-3) -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

    like this or use

    x=\frac{-b\mp \sqrt{b^2-4ac}}{2a} if you do not understand it
    I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
    \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 As I would with  x^2-5x+6 ?

    thanks.
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  14. #14
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    Quote Originally Posted by Tweety View Post

    I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
    \tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0 As I would with  x^2-5x+6 ?

    thanks.
    yes ...

    x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6

    the only difference here is that 3 and 2 can be combined, wheras the 1 and \sqrt{3} cannot.
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  15. #15
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    Quote Originally Posted by skeeter View Post
    yes ...

    x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6

    the only difference here is that 3 and 2 can be combined, wheras the 1 and \sqrt{3} cannot.
    oh right finally I get it, that makes so much sense now!!

    Thanks!
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