How do you solve #3\csc ^ { 2} x + \csc x - 2= 0#?

Question

953 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-22T17:54:03

Multiply both sides of the equation by #sin^2(x)#:

#3+sin(x)-2sin^2(x) = 0#

Multiply both sides by -1:

#2sin^2(x)-sin(x)-3 = 0#

Factor:

#(2sin(x)-3)(sin(x)+1) = 0#

#sin(x) = 3/2# and #sin(x) = -1#

The range for the sine function is #-1<=sin(x)<=1#, therefore, we must discard the first root:

#cancel(sin(x) = 3/2)# and #sin(x) = -1#

Use the inverse sine function n the second root:

#x = sin^-1(-1)#

This is well known to be:

#x = (3pi)/2+2pin; n in ZZ#