# How do you find a numerical value of one trigonometric function of x given cscx/cotx=sqrt2?

Nov 12, 2016

$\therefore x = \pm \frac{\pi}{4} + 2 \pi n$

#### Explanation:

$\csc \frac{x}{\cot} x = \sqrt{2}$

$\frac{\frac{1}{\sin} x}{\cos \frac{x}{\sin} x} = \sqrt{2}$

$\frac{1}{\sin} x \cdot \sin \frac{x}{\cos} x = \sqrt{2}$

$\frac{1}{\cancel{\sin}} x \cdot \cancel{\sin} \frac{x}{\cos} x = \sqrt{2}$

$\frac{1}{\cos} x = \sqrt{2}$

$1 = \sqrt{2} \cos x$

$\frac{1}{\sqrt{2}} = \cos x$

${\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = x$

$\therefore x = \pm \frac{\pi}{4} + 2 \pi n$