How do you solve 4cotx=cotxsin^2x4cotx=cotxsin2x for 0<=x<=2pi0x2π?

1 Answer
Oct 1, 2016

x=pi/2x=π2 and x=(3pi)/2x=3π2

Explanation:

4cotx=cotxsin^2x4cotx=cotxsin2x

hArrcotxsin^2x-4cotx=0cotxsin2x4cotx=0

or cotx(sin^2x-4)=0cotx(sin2x4)=0

As sin^2xsin2x ranges from 00 to 11, sin^2x-4!=0sin2x40

and hence we must have cotx=0cotx=0.

Between 00 and 2pi2π, cotx=0cotx=0 at x=pi/2x=π2 and x=(3pi)/2x=3π2 and hence that is the solution,