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C3: Solving Trigonometric Equations

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    Q. Solve the following equation for 0^o \leq \theta \leq 360^o: \mathrm{cosec^2} \theta = 9\sec^2 \theta

    So far I have:
    1+\cot^2 \theta = 9(1+\tan^2 \theta)
    I don't know where to go from there on.

    Thanks in advance for any help.
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    You may be overcomplicating multiply through by sin^2 or cos^2 at the beginning, this will give you just 1 trig function.

    If I've been ambiguous I apologise, and I'll explain further
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    (Original post by Jozzers)
    Q. Solve the following equation for 0^o \leq \theta \leq 360^o: \mathrm{cosec^2} \theta = 9\sec^2 \theta

    So far I have:
    1+\cot^2 \theta = 9(1+\tan^2 \theta)
    I don't know where to go from there on.

    Thanks in advance for any help.
    Use
    \displaystyle cot \theta =\frac{1}{tan \theta}
    at the LHS so
    \displaystyle \left (1+tan^2\theta \right )\left( \frac{1}{tan^2 \theta}-9\right )=0
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    At the beginning, multiply through by sin^2theta following on from MHRed. It leaves a simple quadratic.
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    (Original post by Jozzers)
    Q. Solve the following equation for 0^o \leq \theta \leq 360^o: \mathrm{cosec^2} \theta = 9\sec^2 \theta

    So far I have:
    1+\cot^2 \theta = 9(1+\tan^2 \theta)
    I don't know where to go from there on.

    Thanks in advance for any help.
     \displaystyle \mathrm{cosec^2} \theta = 9\sec^2 \theta

     \displaystyle \frac{1}{sin^2\theta} = \frac9{cos^2\theta}

    Multiply both sides by  \displaystyle sin^2\theta
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    cosec^2(x) = 9sec^2(x)

    convert cosec onto 1/sin and sec into 1/cos

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

    convert cos^2 into 1-sin^2

    1/sin^2(x) = 9/1-sin^2(x)

    cross multiply

    1-sin^2(x) = 9sin^2(x)

    add sin^2

    1 = 10sin^2(x)

    divide by 10

    1/10 = sin^2(x)

    squareroot

    root(1/10) = sin(x)

    inverse sin

    sin^-1 (root(1/10)) = x

    then take 180 minus your value for second principal value.

    ...hope you can follow this
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    Hahahaha, when you spend ages writing a reply and someone else does it much simpler, cool move Pete xD
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    (Original post by PeteyB26)
    cosec^2(x) = 9sec^2(x)

    convert cosec onto 1/sin and sec into 1/cos

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

    convert cos^2 into 1-sin^2

    1/sin^2(x) = 9/1-sin^2(x)

    cross multiply

    1-sin^2(x) = 9sin^2(x)

    add sin^2

    1 = 10sin^2(x)

    divide by 10

    1/10 = sin^2(x)

    squareroot

    root(1/10) = sin(x)

    inverse sin

    sin^-1 (root(1/10)) = x

    then take 180 minus your value for second principal value.

    ...hope you can follow this
    this is exactly what i was thinking. Just rewriting cosec^2 and sec^2 is the big trick in this question!

    I don't know why people are going for the tan's and cot^s lol
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    (Original post by James A)
    this is exactly what i was thinking. Just rewriting cosec^2 and sec^2 is the big trick in this question!

    I don't know why people are going for the tan's and cot^s lol
    Absolutely, that's the problem with trig identities though, if you go down the wrong path at first you can spend ages just getting yourself in a muddle! They used to scare me so much in C2.. Hahaha
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    (Original post by PeteyB26)
    cosec^2(x) = 9sec^2(x)

    convert cosec onto 1/sin and sec into 1/cos

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

    convert cos^2 into 1-sin^2

    1/sin^2(x) = 9/1-sin^2(x)

    cross multiply

    1-sin^2(x) = 9sin^2(x)

    add sin^2

    1 = 10sin^2(x)

    divide by 10

    1/10 = sin^2(x)

    squareroot

    root(1/10) = sin(x)

    inverse sin

    sin^-1 (root(1/10)) = x

    then take 180 minus your value for second principal value.

    ...hope you can follow this
    Or you can do an even simpler method.

    convert cosec into 1/sin and sec into 1/cos

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

    Cross multiply and you get:

    cos^2(x)/sin^2(x)=9

    1/tan^2(x)=9

    Reciprocate it

    tan^2(x)=1/9

    tan(x)=1/3

    x=18.43,198.43
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    or
     cosec^2\theta = 9sec^2\theta

     \frac{1}{sin^2\theta} = \frac{9}{cos^2\theta}

     1 = \frac{9sin^2\theta}{cos^2\theta}

     1 = 9tan^2\theta

     \frac{1}{9} = tan^2\theta

     \pm(\frac{1}{3}) = tan\theta

    edit: typically someone beats me to it
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    (Original post by PeteyB26)
    Absolutely, that's the problem with trig identities though, if you go down the wrong path at first you can spend ages just getting yourself in a muddle! They used to scare me so much in C2.. Hahaha
    yeah i know, it helps alot if i had a brain that could see five steps in advance haha
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    Ohhh I get it now.

    Thank you everyone!

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Updated: April 3, 2012
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