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

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Answer 1 · 2018-03-19T08:20:28

$θ = {cos}^{- 1} (\frac{1}{\sqrt{\sqrt{2}}}), 2 π - {cos}^{- 1} (\frac{1}{\sqrt{\sqrt{2}}})$ $theta=cos^-1(1/sqrtsqrt2), 2pi-cos^-1(1/sqrtsqrt2)$

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

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

$2 {(\frac{1}{sin x})}^{2} - {(\frac{cos x}{sin x})}^{4} = - 1$ $2(1/sinx)^2-(cosx/sinx)^4=-1$

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

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

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

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

${cos}^{4} x + {sin}^{4} x = {({cos}^{2} x + {sin}^{2} x)}^{2} - 2 {cos}^{2} x {sin}^{2} x$ $cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x$

$2 {sin}^{2} x = {({cos}^{2} x + {sin}^{2} x)}^{2} - 2 {cos}^{2} x {sin}^{2} x$ $2sin^2x=(cos^2x+sin^2x)^2-2cos^2xsin^2x$

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

$2 {sin}^{2} x = {(1)}^{2} - 2 {cos}^{2} x {sin}^{2} x$ $2sin^2x=(1)^2-2cos^2xsin^2x$

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

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

Substituting

$2 (1 - u) = 1 - 2 u (1 - u)$ $2(1-u)=1-2u(1-u)$

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

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

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

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

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

$u = \pm (\frac{1}{\sqrt{2}})$ $u=+-(1/sqrt2)$

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

${cos}^{2} x = \pm (\frac{1}{\sqrt{2}})$ $cos^2x=+-(1/sqrt2)$

Considering only positive value for real numbers

${cos}^{2} x = \frac{1}{\sqrt{2}}$ $cos^2x=1/sqrt2$

$cos x = \pm (\frac{1}{\sqrt{\sqrt{2}}})$ $cosx=+-(1/sqrtsqrt2)$

$θ = {cos}^{- 1} (\frac{1}{\sqrt{\sqrt{2}}}), 2 π - {cos}^{- 1} (\frac{1}{\sqrt{\sqrt{2}}})$ $theta=cos^-1(1/sqrtsqrt2), 2pi-cos^-1(1/sqrtsqrt2)$