How do you solve #csc^2x-cscx+9=11# for #0<=x<=2pi#?

Question

4397 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-19T14:55:50

#x = (pi) / (6), (5 pi) / (6), (3 pi) / (2)#

We have: #csc^(2)(x) - csc(x) + 9 = 11#; #0 leq x leq 2 pi#

#=> csc^(2)(x) - csc(x) - 2 = 0#

#=> csc^(2)(x) + csc(x) - 2 csc(x) - 2 = 0#

#=> csc(x) (csc(x) + 1) - 2 (csc(x) + 1) = 0#

#=> (csc(x) + 1 ) (csc(x) - 2) = 0#

#=> csc(x) + 1 = 0#

#=> csc(x) = - 1#

#=> x = csc^(- 1)(- 1)#

#=> x = (3 pi) / (2)#

or

#=> csc(x) - 2 = 0#

#=> csc(x) = 2#

#=> x = csc^(- 1)(2)#

#=> x = (pi) / (6)#

Then, the domain given is #0 leq x leq 2 pi#:

#=> x = (pi) / (6), (3 pi) / (2), pi - (pi) / (6)#

#=> x = (pi) / (6), (5 pi) / (6), (3 pi) / (2)#