Please help me to solve x , 0 <= x <= 360, where
3 tan x - 2 + 3 sec x = 2 cosec x
Thank you
$\displaystyle 3 tan(x) - 2 + 3 sec(x) = 2 csc(x)$
$\displaystyle 3 \frac{sin(x)}{cos(x)} - 2 + 3 \frac{1}{cos(x)} = 2 \frac{1}{sin(x)}$ <-- 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