# How do you find the general solutions for cot^2 x +csc x = 1?

Aug 10, 2015

$x = n \pi - \frac{\pi}{6} {\left(- 1\right)}^{n} \text{ or " x = 2npi + pi/2 ", } n \in \mathbb{Z}$

#### Explanation:

${\sin}^{2} x + {\cos}^{2} x = 1$
Divide by ${\sin}^{2} x$ gives: $1 + {\cot}^{2} x = {\csc}^{2} x$

Therefore: ${\csc}^{2} x + \csc x - 2 = 0$
$\left(\csc x + 2\right) \left(\csc x - 1\right) = 0$
$\frac{1}{\sin} x = - 2 \text{ or } \frac{1}{\sin} x = 1$
$\sin x = - \frac{1}{2} \text{ or } \sin x = 1$
$x = n \pi - \frac{\pi}{6} {\left(- 1\right)}^{n} \text{ or " x = 2npi + pi/2 ", } n \in \mathbb{Z}$