# How do you solve this equation: csc^4 2(u)-4=0 ?

## i use the inverse that is sin?

May 20, 2018

Given ${\csc}^{4} \left(2 u\right) - 4 = 0$

Factor as ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ where $a = {\cos}^{2} \left(2 u\right)$ and $b = 4$

$\left({\csc}^{2} \left(2 u\right) - 2\right) \left({\csc}^{2} \left(2 u\right) + 2\right) = 0$

To find the values where the above equation is true, we must set both factors equal to 0:

${\csc}^{2} \left(2 u\right) - 2 = 0 \mathmr{and} {\csc}^{2} \left(2 u\right) + 2 = 0$

Substitute ${\csc}^{2} \left(2 u\right) = \frac{1}{\sin} ^ 2 \left(2 u\right)$

$\frac{1}{\sin} ^ 2 \left(2 u\right) - 2 = 0 \mathmr{and} \frac{1}{\sin} ^ 2 \left(2 u\right) + 2 = 0$

Add two to both sides of the first equation and subtract 2 from both sides of the second equation:

$\frac{1}{\sin} ^ 2 \left(2 u\right) = 2 \mathmr{and} \frac{1}{\sin} ^ 2 \left(2 u\right) = - 2$

Invert both sides of both equations:

${\sin}^{2} \left(2 u\right) = \frac{1}{2} \mathmr{and} {\sin}^{2} \left(2 u\right) = - \frac{1}{2}$

Take the square root of both sides of both equations:

$\sin \left(2 u\right) = \pm \frac{\sqrt{2}}{2} \mathmr{and} \sin \left(2 u\right) = \pm \frac{\sqrt{2}}{2} i$

Take the inverse sine of both sides:

$2 u = \pm {\sin}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) \mathmr{and} 2 u = \pm \sin \left(\frac{\sqrt{2}}{2} i\right)$

The first equation is well known but the second equation becomes the inverse hyperbolic sine:

$2 u = \pm \frac{\pi}{4} \mathmr{and} 2 u = \pm i {\sinh}^{-} 1 \left(\frac{\sqrt{2}}{2}\right)$

It will be periodic at integer multiples of $\pi$"

$2 u = n \pi \pm \frac{\pi}{4} \mathmr{and} 2 u = n \pi \pm i {\sinh}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) , n \in \mathbb{Z}$

Divide by 2:

$u = n \frac{\pi}{2} \pm \frac{\pi}{8} \mathmr{and} u = n \frac{\pi}{2} \pm i \frac{1}{2} {\sinh}^{-} 1 \left(\frac{\sqrt{2}}{2}\right)$

May 20, 2018

pi/8 + kpi; and (3pi)/8 + kpi
(5pi)/8 + kpi; and (7pi)/8 + kpi

#### Explanation:

$\csc u = \frac{1}{\sin} u$
$\frac{1}{{\sin}^{4} 2 u} = 4$
$\frac{1}{\sin} ^ 2 2 u = 2$
${\sin}^{2} 2 u = \frac{1}{2}$
$\sin 2 u = \pm \frac{1}{\sqrt{2}}$
Trig table and unit circle give 4 solutions:
1. $\sin 2 u = \frac{1}{\sqrt{2}}$ -->
$2 u = \frac{\pi}{4}$ and $2 u = \frac{3 \pi}{4}$
a. $2 u = \frac{\pi}{4} + 2 k \pi$ --> $u = \frac{\pi}{8} + k \pi$
b. $2 u = \frac{3 \pi}{4} + 2 k \pi$--> $u = \frac{3 \pi}{8} + k \pi$
2. $\sin 2 u = - \frac{1}{\sqrt{2}}$ -->
$2 u = - \frac{\pi}{4}$, or $2 u = \frac{7 \pi}{4}$, (co-terminal), and
$2 u = \pi - \left(- \frac{\pi}{4}\right) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$
a. $2 u = \frac{5 \pi}{4}$ --> $u = \frac{5 \pi}{8} + k \pi$
b. $2 u = \frac{7 \pi}{4} + 2 k \pi$ --> $u = \frac{7 \pi}{8} + k \pi$