# need help with a trig problem

• March 15th 2009, 04:11 PM
RaphaelB30
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
• March 15th 2009, 04:24 PM
e^(i*pi)
Quote:

Originally Posted by RaphaelB30
this is the problem: 2 csc²x=3 cot²x-1
this is what I have so far: see attached file

$\frac{2}{sin^2(x)} = \frac{3cos^2(x)}{sin^2(x)} - 1$

Getting the same denominator on the right:

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

Now $sin^2(x)$ will cancel

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

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

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

using the difference of two squares:

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

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

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

$x = arcos(-\frac{\sqrt3}{2}) = \frac{\pi}{6}$
• March 15th 2009, 05:31 PM
skeeter
Quote:

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

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

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

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

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

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

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