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

i use the inverse that is sin?

2 Answers
May 20, 2018

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)#

May 20, 2018

#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#