Question

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

i use the inverse that is sin?

Answer 1

Given `csc^4(2u) -4 = 0`

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

`(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(2u) -2 = 0 and csc^2(2u)+2 = 0`

Substitute `csc^2(2u) = 1/sin^2(2u)`

`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:

`1/sin^2(2u) =2 and 1/sin^2(2u)=-2`

Invert both sides of both equations:

`sin^2(2u) =1/2 and sin^2(2u)=-1/2`

Take the square root of both sides of both equations:

`sin(2u) =+-sqrt2/2 and sin(2u)=+-sqrt2/2i`

Take the inverse sine of both sides:

`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:

`2u =+-pi/4 and 2u=+-isinh^-1(sqrt2/2)`

It will be periodic at integer multiples of `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

`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`