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

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

i use the inverse that is sin?

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Answer 1 · 2018-05-20T21:26:01

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

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

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

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

${csc}^{2} (2 u) - 2 = 0 and {csc}^{2} (2 u) + 2 = 0$ $csc^2(2u) -2 = 0 and csc^2(2u)+2 = 0$

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

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

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

$\frac{1}{{sin}^{2} (2 u)} = 2 and \frac{1}{{sin}^{2} (2 u)} = - 2$ $1/sin^2(2u) =2 and 1/sin^2(2u)=-2$

Invert both sides of both equations:

${sin}^{2} (2 u) = \frac{1}{2} and {sin}^{2} (2 u) = - \frac{1}{2}$ $sin^2(2u) =1/2 and sin^2(2u)=-1/2$

Take the square root of both sides of both equations:

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

Take the inverse sine of both sides:

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

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

$2 u = \pm \frac{π}{4} and 2 u = \pm i {sinh}^{- 1} (\frac{\sqrt{2}}{2})$ $2u =+-pi/4 and 2u=+-isinh^-1(sqrt2/2)$

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

$2u =npi +-pi/4 and 2u= npi+-isinh^-1(sqrt2/2), n in ZZ$

Divide by 2:

$u =npi/2+-pi/8 and u=npi/2+-i1/2sinh^-1(sqrt2/2)$

Answer 2 · 2018-05-20T21:32:48

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

Explanation:

$csc u = 1/sin u$
$1/(sin^4 2u) = 4$
$1/sin^2 2u = 2$
$sin^2 2u = 1/2$
$sin 2u = +- 1/sqrt2$
Trig table and unit circle give 4 solutions:
1. $sin 2u = 1/sqrt2$ -->
$2u = pi/4$ and $2u = (3pi)/4$
a. $2u = pi/4 + 2kpi$ --> $u = pi/8 + kpi$
b. $2u = (3pi)/4 + 2kpi$ --> $u = (3pi)/8 + kpi$
2. $sin 2u = - 1/sqrt2$ -->
$2u = - pi/4$ , or $2u = (7pi)/4$ , (co-terminal), and
$2u = pi - (-pi/4) = pi + pi/4 = (5pi)/4$
a. $2u = (5pi)/4$ --> $u = (5pi)/8 + kpi$
b. $2u = (7pi)/4 + 2kpi$ --> $u = (7pi)/8 + kpi$

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

i use the inverse that is sin?

2 Answers

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