# Solve the equation for 0<x<360 2csc^2x-cot^4x=-1?

Mar 19, 2018

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

#### Explanation:

Given:
$2 {\csc}^{2} x - {\cot}^{4} x = - 1$
$\csc x = \frac{1}{\sin} x$

$\cot x = \cos \frac{x}{\sin} x$

$2 {\left(\frac{1}{\sin} x\right)}^{2} - {\left(\cos \frac{x}{\sin} x\right)}^{4} = - 1$

$\frac{2}{\sin} ^ 2 x - {\cos}^{4} \frac{x}{\sin} ^ 4 x = - 1$

Multiplying throughout by ${\sin}^{4} x$

$2 {\sin}^{2} x - {\cos}^{4} x = - {\sin}^{4} x$
Transposing

$2 {\sin}^{2} x = {\cos}^{4} x + {\sin}^{4} x$

${\cos}^{4} x + {\sin}^{4} x = {\left({\cos}^{2} x + {\sin}^{2} x\right)}^{2} - 2 {\cos}^{2} x {\sin}^{2} x$

$2 {\sin}^{2} x = {\left({\cos}^{2} x + {\sin}^{2} x\right)}^{2} - 2 {\cos}^{2} x {\sin}^{2} x$

${\cos}^{2} x + {\sin}^{2} x = 1$

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

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

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

Substituting

$2 \left(1 - u\right) = 1 - 2 u \left(1 - u\right)$

$2 - 2 u = 1 - 2 u + 2 {u}^{2}$

$2 {u}^{2} - 2 u + 2 u + 1 - 2 = 0$

$2 {u}^{2} - 1 = 0$

$2 {u}^{2} = 1$

${u}^{2} = \frac{1}{2}$

$u = \pm \left(\frac{1}{\sqrt{2}}\right)$

$u = {\cos}^{2} x$

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

Considering only positive value for real numbers

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

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

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