# Thread: need help with a trig problem

1. ## need help with a trig problem

this is the problem: 2 csc²x=3 cot²x-1
this is what I have so far: see attached file

2. Originally Posted by RaphaelB30
this is the problem: 2 csc²x=3 cot²x-1
this is what I have so far: see attached file
$\displaystyle \frac{2}{sin^2(x)} = \frac{3cos^2(x)}{sin^2(x)} - 1$

Getting the same denominator on the right:

$\displaystyle \frac{3cos^2(x) - sin^2(x)}{sin^2(x)}$

Now $\displaystyle sin^2(x)$ will cancel

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

$\displaystyle 2 = 3cos^2(x) - (1-cos^2(x)) = 3cos^2(x) - 1 + cos^2(x) = 4cos^2(x) - 1$

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

using the difference of two squares:

$\displaystyle (2cos(x) - \sqrt3)(2cos(x) + \sqrt3) = 0$

so $\displaystyle cos(x) = ±\frac{\sqrt3}{2}$ and thus

$\displaystyle x = arcos(\frac{\sqrt3}{2}) = \frac{5\pi}{6}$ or

$\displaystyle x = arcos(-\frac{\sqrt3}{2}) = \frac{\pi}{6}$

3. 2 csc²x=3 cot²x-1
$\displaystyle 2(\cot^2{x} + 1) = 3\cot^2{x} - 1$

$\displaystyle 0 = \cot^2{x} - 3$

$\displaystyle 0 = (\cot{x} - \sqrt{3})(\cot{x} + \sqrt{3})$

$\displaystyle \cot{x} = \sqrt{3}$

$\displaystyle x = \frac{\pi}{6}$ , $\displaystyle x = \frac{7\pi}{6}$

$\displaystyle \cot{x} = -\sqrt{3}$

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