Question #ef674

1 Answer
Feb 15, 2018

#color(blue)(pi/4 , (3pi)/4)#

Explanation:

#2csc^2(2x)-3csc(2x)+1=0#

Let:

#u=csc(2x)#

Then:

#2u^2-3u+1=0#

Factor:

#(2u-1)(u-1)=0=> u=1 and u=1/2#

But #u=csc2x=>csc2x=1 and csc2x=1/2#

#csc2x=1=>2x=csc^-1(1)=pi/2=>x=pi/4,(3pi)/4#

#csc2x=1/2=>2x=csc^-1(1/2)#= no real solutions. *

#csc2x=1/(sin2x)#

  • #csc2x=1/2=>sin2x=2color(white)(88)# not possible for real values.

#-1<=sinx<=1#

So our solutions are only.

#color(blue)(pi/4 , (3pi)/4)#