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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- Jun 19th 2007, 11:38 PMNicoSolve 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 - Jun 20th 2007, 04:15 AMtopsquark
$\displaystyle 3 tan(x) - 2 + 3 sec(x) = 2 csc(x)$

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

$\displaystyle 3 sin^2(x) - 2 sin(x)cos(x) + 3 sin(x) = 2 cos(x)$

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

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

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

$\displaystyle 3 sin^2(x) + 3 sin(x) = 2(sin(x) + 1) cos(x)$ <-- Square both sides

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

Now $\displaystyle cos^2(x) = 1 - sin^2(x)$. (If we hadn't squared the equation we'd've had to choose which quadrant x is in.)

$\displaystyle 9 sin^4(x) + 18sin^3(x) + 9 sin^2(x) = (4sin^2(x) + 8sin(x) + 4)(1 - sin^2(x))$

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

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

Now let $\displaystyle y = sin(x)$

$\displaystyle 13y^4 + 26y^3 + 9y^2 - 8y - 4 = 0$

This is ugly. BUT, if we use the rational root theorem we know that all the possible rational roots are at $\displaystyle y = \pm 1, \pm \frac{2}{13}, \pm \frac{4}{13}$. It turns out that $\displaystyle y = -1$ is a rational zero. So by division we get that

$\displaystyle 13y^4 + 26y^3 + 9y^2 - 8y - 4 = (y + 1)(13y^3 + 13y^2 - 4y - 4) = 0$

We may apply the rational root theorem again to find that $\displaystyle y = -1$ is again a zero of the cubic polynomial, or you may note that we may factor the cubic polynomial by grouping:

$\displaystyle 13y^3 + 13y^2 - 4y - 4 = (13y^3 + 13y^2) + (-4y - 4)$

$\displaystyle = 13y^2(y + 1) - 4(y + 1) = (13y^2 - 4)(y + 1)$

So:

$\displaystyle 13y^4 + 26y^3 + 9y^2 - 8y - 4 = (y + 1)^2(13y^2 - 4) = 0$

So we have that

$\displaystyle y = 1$

and

$\displaystyle y = \pm \frac{2}{\sqrt{13}}$

The $\displaystyle y = 1$ solution generates

$\displaystyle sin(x) = 1$

$\displaystyle x = 90^o$

The other two are not so nice. We obtain:

$\displaystyle sin(x) = \frac{2}{\sqrt{13}}$

which has only the approximate solutions:

$\displaystyle x \approx 33.6901^o, 146.3099^o$

and

$\displaystyle sin(x) = -\frac{2}{\sqrt{13}}$

which has only the approximate solutions:

$\displaystyle x \approx 213.6901^o, 326.3099^o$

-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 $\displaystyle x \approx 33.6901^o, ~ 213.6901^o$ solutions work.

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

-Dan - Jun 20th 2007, 04:29 AMNico
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 - Jun 20th 2007, 04:32 AMtopsquark