What is the general solution of trigonometric equation #sqrt3cotx+1=0#?

Question

6603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T07:14:25

General solution of #sqrt3cotx+1=0# is #x=npi+(2pi)/3#, where #n# is an integer.

As #sqrt3cotx+1=0#, we have

#sqrt3cotx=-1#

or #cotx=-1/sqrt3#

As #cot(pi/3)=1/sqrt3# and #cot(pi-pi/3)=1/sqrt3#

i.e. #cotx=cot((2pi)/3)#

Now as #cot# function has a cycle of #pi# radians or #180^o#, it repeats after every #pi# radians or #180^o#.

General solution of #sqrt3cotx+1=0# is #x=npi+(2pi)/3#, where #n# is an integer.