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    Dear All,
    I have already sweated over this question for an hour and its starting to get on my nerves. I was hoping if the brightest minds on TSR would mind giving me a helping hand.
    Solve in the interval 0< \theta<180^{\circ}
    2 \cot^2 \theta=7 \mathrm{cosec} \theta-8
    Thanks,
    Saran
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    change the cot into cosec and form a quadratic, bam!

    2cot^2=2(cosec^2-1)
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    1+Cot^2theta=cosec^2theta
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    (Original post by Diaz)
    1+Cot^2theta=sec^2theta
    Aint it cosec?
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    (Original post by Diaz)
    1+Cot^2theta=sec^2theta
    :nah: It's cosecant:

    \displaystyle 1 + \cot^2(\theta) = \csc^2(\theta)
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    (Original post by Wesssty)
    Aint it cosec?
    What happens when you do it in your head:rolleyes:
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    (Original post by saran35)
    Dear All,
    I have already sweated over this question for an hour and its starting to get on my nerves. I was hoping if the brightest minds on TSR would mind giving me a helping hand.
    2 \cot^2 \theta=7 \mathrm{cosec} \theta-8
    Thanks,
    Saran
    cot^2(theta) = cosec^2(theta) - 1

    so 2cosec^2theta - 2 = 7cosec theta - 8
    assume cosec theta = y
    2y^2 - 2 =7y-8
    2y^2 -7y + 6 = 0
    (2y-3)(y-2) = 0
    y = 3/2 or 2
    so 1/sin^2(theta) = 3/2 or 2

    sin^2(theta) = 2/3 or 1/2
    then square root everything and solve normally.
 
 
 
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