# Thread: [SOLVED] solving an equation

1. ## [SOLVED] solving an equation

I seem to have partially gotten this right, but am missing some answers, so chances are i stumbled across the (partial) right answer improperly. If someone could tell me what I'm doing wrong, that would be grand

on an interval of 0 - 2pi
$tan2x - cotx = 0$
using double angle identities and cotanget identity
$\frac{2tanx}{1-tan^2x} - \frac{1}{tanx} = 0$
$\frac{2tanx}{1-tan^2x} = \frac{1}{tanx}$
multiply both sides by tanx
$\frac{2tan^2x}{1-tan^2x} = 1$
multiply both sides by $1-tan^2x$
$2tan^2x = 1-tan^2x$
$3tan^2x = 1$
$tan^x = \frac{1}{3}$
$tanx = \sqrt{\frac{1}{3}}$

so that gives an answer at $\frac{pi}{6}, \frac{5pi}{6}, \frac{7pi}{6} and \frac{11pi}{6}$, but the book specifies additional answers at $\frac{pi}{2}, \frac{3pi}{2}$. Did I make a mistake above? If not, where are those 2 additional answers from?

2. I think you "lost" the two additional solutions when you multiplied by $\tan{x}$ .

$\tan(2x) - \cot{x} = 0$

$\frac{\sin(2x)}{\cos(2x)} - \frac{\cos{x}}{\sin{x}} = 0
$

$\frac{2\sin{x}\cos{x}}{2\cos^2{x}-1} - \frac{\cos{x}}{\sin{x}} = 0$

$\frac{2\sin^2{x}\cos{x}}{\sin{x}(2\cos^2{x}-1)} - \frac{\cos{x}(2\cos^2{x}-1)}{\sin{x}(2\cos^2{x}-1)} = 0$

$2\sin^2{x}\cos{x} - \cos{x}(2\cos^2{x}-1) = 0
$

$2(1-\cos^2{x})\cos{x} - \cos{x}(2\cos^2{x}-1) = 0
$

$2\cos{x} - 2\cos^3{x} - 2\cos^3{x} + \cos{x} = 0$

$3\cos{x} - 4\cos^3{x} = 0$

$\cos{x}(3 - 4\cos^2{x}) = 0$

$\cos{x} = 0$

$x = \frac{\pi}{2} \, , \, \frac{3\pi}{2}$

$\cos{x} = \pm \frac{\sqrt{3}}{2}$

$x = \frac{\pi}{6} \, , \, \frac{5\pi}{6} \, , \, \frac{7\pi}{6} \, , \, \frac{11\pi}{6}$

3. ah, that's right, you're not supposed to multiply/divide by trig functions for fear of losing a solution. Totally forgot that. Thanks for the assistance.