# How do I solve for 0 ≤ x < 2π using the equation cot² x - csc x =1 ??

Feb 22, 2018

$x = \frac{\pi}{4} , \frac{3}{4} \pi \mathmr{and} \frac{3}{2} \pi$

#### Explanation:

${\cot}^{2} x - \csc x = 1$

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

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

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

${\cos}^{2} x - \sin x = {\sin}^{2} x$

change ${\cos}^{2} x \to 1 - {\sin}^{2} x$
$1 - {\sin}^{2} x - \sin x = {\sin}^{2} x$

rearrange the equation
$0 = 2 {\sin}^{2} x + \sin x - 1$

factorize the equation
$0 = \left(2 \sin x - 1\right) \left(\sin x + 1\right)$

therefore,
$2 \sin x - 1 = 0 \mathmr{and} \sin x + 1 = 0$
#sin x = 1/2 and sin x = -1

$x = \frac{\pi}{4} , \frac{3}{4} \pi$ and $x = \frac{3}{2} \pi$