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    Solve in the interval 0< (theta) < 180 degrees

    2cot^2(theta)= 7cosec(theta) - 8

    Confused on how to answer this.

    I made the 2 cot^2 theta 2cos^2(theta)/sin^2(theta) and the 7cosec(theta) 7/sin(theta) but now I don't know what to do many thanks
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    Multiply through by \sin^{2}\theta and then replace the cos with the equivalent in sin.
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    I still confused, So I've times by sin^2 (theta) so I've got

    2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

    So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

    so 7sin(theta) = 10 which doesnt work? What have I done wrong?
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    I still confused, So I've times by sin^2 (theta) so I've got

    2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

    So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

    so 7sin(theta) = 10 which doesnt work? What have I done wrong?
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    (Original post by joestevens2092)
    I still confused, So I've times by sin^2 (theta) so I've got

    2cos^2(theta)=7sin(theta) - 8 sin^2(theta)

    So then 8sin^2(theta)+2cos^2(theta) = 7sin(theta)

    so 7sin(theta) = 10 which doesnt work? What have I done wrong?
    Ah, I assume you've tried to use \sin^{2}\theta + \cos^{2}\theta = 1??

    Unfortunately that only works when the coefficients are the same (obviously you'd need to change the constant to match the coefficients).

    Re-write the cos^2 in terms of sin and then solve as you should be left with a quadratic in sin.
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    got it cheers
 
 
 
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