This is a trigonometric equation I can't figure out? #cot^2x = -2cotx -1#

1 Answer
Apr 27, 2018

#x=(2k+1)pi-pi/4,kinZZ#

Explanation:

Here,

#Cot^2x = -2cotx -1 #

#=>cot^2x+2cotx+1=0#

#=>(cotx+1)^2=0#

#=>cotx+1=0#

#color(red)(=>cotx=-1...(A)#

#=>tanx=-1...# #tocosx!=0#

#=>tanx=tan(-pi/4)#

#color(blue)(=>x=kpi-pi/4,kinZZ...to(B)#

We know that ,the range of #cot^-1x, is :(0.pi)#

Now from #(A)#

#cotx=-1=>x=cot^-1(-1)!=-pi/4...toIV^(th)Quadrant#

#and-pi/4!in(0,pi)#

So, #color(red)(cot^-1(-1)=pi-pi/4=(3pi)/4...toII^(nd)Quadrant#

Hence, from #(B)#

#x=(2k+1)pi-pi/4,kinZZ#

Note:

#color(blue)(x={color(red)(kpi)-pi/4,kinZZ}#

#:.x={color(red)((2k+1)pi)-pi/4,kinZZ}uu{color(red)((2k)pi)- pi/4,kinZZ}#
#color(white)(.................)color(red)(II^(nd)Quadrant)color(white) (...................)color(red)(IV^(th)Quadrant)#