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Math Help - Solve x (tan sec and cosec)

  1. #1
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    Solve x (tan sec and cosec)

    Please help me to solve x , 0 <= x <= 360, where

    3 tan x - 2 + 3 sec x = 2 cosec x

    Thank you
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Nico View Post
    Please help me to solve x , 0 <= x <= 360, where

    3 tan x - 2 + 3 sec x = 2 cosec x

    Thank you
    3 tan(x) - 2 + 3 sec(x) = 2 csc(x)

    3 \frac{sin(x)}{cos(x)} - 2 + 3 \frac{1}{cos(x)} = 2 \frac{1}{sin(x)} <-- Clear the fractions

    3 sin^2(x) - 2 sin(x)cos(x) + 3 sin(x) = 2 cos(x)

    3 sin^2(x) - 2 sin(x)cos(x) + 3 sin(x) - 2 cos(x) = 0

    3 sin^2(x) + 3 sin(x) - 2(sin(x) + 1)cos(x) = 0

    We want to get this equation in terms of a single trig function. We can do that by employing sin^2(x) + cos^2(x) = 1, so we need to do the following:
    3 sin^2(x) + 3 sin(x) =  2(sin(x) + 1) cos(x) <-- Square both sides

    9 sin^4(x) + 18sin^3(x) + 9 sin^2(x) = (4sin^2(x) + 8sin(x) + 4)cos^2(x)

    Now cos^2(x) = 1 - sin^2(x). (If we hadn't squared the equation we'd've had to choose which quadrant x is in.)
    9 sin^4(x) + 18sin^3(x) + 9 sin^2(x) = (4sin^2(x) + 8sin(x) + 4)(1 - sin^2(x))

    9 sin^4(x) + 18sin^3(x) + 9 sin^2(x) = 4sin^2(x) + 8sin(x) + 4 - 4sin^4(x) - 8sin^3(x) - 4sin^2(x)

    13 sin^4(x) + 26sin^3(x) + 9 sin^2(x) - 8sin(x) - 4 = 0

    Now let y = sin(x)
    13y^4 + 26y^3 + 9y^2 - 8y - 4 = 0

    This is ugly. BUT, if we use the rational root theorem we know that all the possible rational roots are at y = \pm 1, \pm \frac{2}{13}, \pm \frac{4}{13}. It turns out that y = -1 is a rational zero. So by division we get that
    13y^4 + 26y^3 + 9y^2 - 8y - 4 = (y + 1)(13y^3 + 13y^2 - 4y - 4) = 0

    We may apply the rational root theorem again to find that y = -1 is again a zero of the cubic polynomial, or you may note that we may factor the cubic polynomial by grouping:
    13y^3 + 13y^2 - 4y - 4 = (13y^3 + 13y^2) + (-4y - 4)

    = 13y^2(y + 1) - 4(y + 1) = (13y^2 - 4)(y + 1)

    So:
    13y^4 + 26y^3 + 9y^2 - 8y - 4 = (y + 1)^2(13y^2 - 4) = 0

    So we have that
    y = 1
    and
    y = \pm \frac{2}{\sqrt{13}}

    The y = 1 solution generates
    sin(x) = 1

    x = 90^o

    The other two are not so nice. We obtain:
    sin(x) = \frac{2}{\sqrt{13}}
    which has only the approximate solutions:
    x \approx 33.6901^o, 146.3099^o

    and
    sin(x) = -\frac{2}{\sqrt{13}}
    which has only the approximate solutions:
    x \approx 213.6901^o, 326.3099^o

    -Dan

    PS Whoops! I almost forgot. We need to check each of these solutions in the original equation to be sure that they ARE actually solutions. I get that only the x \approx 33.6901^o, ~ 213.6901^o solutions work.

    Graphically solving the problem makes it appear that x = 270^o is also a solution. However this x value can't be a solution because sec(180) is undefined.

    -Dan
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  3. #3
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    Thank you, what about:

    3 sin x / cos x - 2 + 3 /cos x = 2 / sin x

    2 sin x.sin x - 2 sin x cos x = 3 sin x = 2 cos x

    3 sinx (sin x +1) = 2 cos x (sin x +1)

    3 sinx = 2 cos x

    tan x = 2/3

    x = 33,69 and 213,69
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Nico View Post
    Thank you, what about:

    3 sin x / cos x - 2 + 3 /cos x = 2 / sin x

    2 sin x.sin x - 2 sin x cos x = 3 sin x = 2 cos x

    3 sinx (sin x +1) = 2 cos x (sin x +1)

    3 sinx = 2 cos x

    tan x = 2/3

    x = 33,69 and 213,69
    Something I've always told my students at the beginning of the semester. I tend to miss the easy way to do the problem! Good job!

    -Dan
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