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C3-trig

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    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
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    (Original post by otrivine)
    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
    You could write sec in terms of cos
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    (Original post by otrivine)
    hi can someone solve this
    3secsquaredX-4=0

    can someone help me
     \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

    Remember,  \displaystyle sec^2x = \frac1{cos^2x}
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    3/cos^2 (x) = 4
    cos^2(x) = 3/4
    cos (x) = root3/2
    x = pi/6 or 30 degrees
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    (Original post by r_t)
    3/cos^2 (x) = 4
    cos^2(x) = 3/4
    cos (x) = root3/2
    x = pi/6 or 30 degrees
    It will be better to give OP some hints, so that he does it himself, rather than post a full solution for him.

    By the way, your solution isn't completely correct.
     \displaystyle cos^2x=\frac34 \implies cosx= \pm \frac{\sqrt3}{2}

    You haven't considered the negative case.
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    (Original post by raheem94)
     \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4

    Remember,  \displaystyle sec^2x = \frac1{cos^2x}
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
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    (Original post by otrivine)
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
    secX does not equal 3/4
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    (Original post by otrivine)
    i get it so secX=3/4 and you invert it and change the root to +4/3 then do cos-1(root3/2)
    I am not understanding what you are trying to do.

      \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

    Do you get it?
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    (Original post by raheem94)
    I am not understanding what you are trying to do.

      \displaystyle 3sec^2x-4=0 \implies 3sec^2x=4 \implies sec^2x=\frac43 \implies \frac1{cos^2x} = \frac43 \\ \implies cos^2x = \frac34

    Do you get it?
    ohhh yes yes i get it now
    and one more thing for this question tansquared2X-2sec2x+1=0
    i got 90 degrees and one maths error is this correct so far or not?
    as for the equation i got secsquared2x-2sec2x+1=0
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    (Original post by otrivine)
    ohhh yes yes i get it now
    and one more thing for this question tansquared2X-2sec2x+1=0
    i got 90 degrees and one maths error is this correct so far or not?
    as for the equation i got secsquared2x-2sec2x+1=0
    It isn't correct.

     tan^22x-2sec2x+1=0

    Remember,  tan^22x=sec^22x-1

    So the equation becomes,
     tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0
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    (Original post by raheem94)
    It isn't correct.

     tan^22x-2sec2x+1=0

    Remember,  tan^22x=sec^22x-1

    So the equation becomes,
     tan^22x-2sec2x+1=0 \implies sec^22x-1-2sec2x+1=0 \\ \implies sec^22x-2sec2x=0
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
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    (Original post by otrivine)
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
    Factorise
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    (Original post by otrivine)
    yes that is what i got but my answer are wrong

    i applied quadratic formula X2-2X+0 which gives 2 and 0
    You don't need to apply the quadratic formula,

     \displaystyle sec^22x-2sec2x=0  \implies sec2x(sec2x-2)=0

    So you get 2 equations,  sec2x=0 \text{ and } sec2x-2 =0
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    (Original post by steve2005)
    Factorise
    yes i get same answer 0 , 2
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    (Original post by raheem94)
    You don't need to apply the quadratic formula,

     \displaystyle sec^22x-2sec2x=0  \implies sec2x(sec2x-2)=0

    So you get 2 equations,  sec2x=0 \text{ and } sec2x-2 =0
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
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    (Original post by otrivine)
    yes i get same answer 0 , 2
    See my previous post.

    0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

    But you need to use  \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'
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    (Original post by otrivine)
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
    Are the answers in your book 0 and 180?
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    (Original post by raheem94)
    See my previous post.

    0 and 2 are correct because the 2 equation are sec2x=0 and sec2x=2.

    But you need to use  \displaystyle sec2x= \frac1{cos2x} to find the values of '2x' and hence 'x'
    ok so dont we have to do firstly cos-1(0) then divide by 2 ?
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    (Original post by raheem94)
    Are the answers in your book 0 and 180?
    yes its the edexcel c3 text book
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    (Original post by otrivine)
    i get 90 degrees and maths error

    and then i takeaway 90 with 360 then keep dividing by 2 correct?

    cause my book i giving me only 2 solution 0,180
    I think you are looking at the wrong answers.

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