It has four solutions.

 Plz give the solution...............

|cot(x)| = cot(x)  + 1/sin(x)

|cot(x)| = cot(x)  + cosec(x)

Now, clearly note that |cot(x)|  = cot(x) , when x belongs to 0 to pi/2 or x belongs from pi to 3pi/2

                                                     = - cot(x), when x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi

CASE1. When x belongs to 0 to pi/2 or x belongs from pi to 3pi/2,

cot(x) = cot(x) + cosec(x)

or, cosec(x) = 0 which is not possible.

CASE2. When x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi

-cot(x) = cot(x) + cosec(x)

(cosec(x))(1 + 2 cos(x)) = 0

cosec(x) # 0

so, 1 + 2cos(x) = 0

or, cos(x) = -1/2

so, x = 2pi/3 (because x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi)

Hence there is only one solution.

You can see it from the graph even---

clearly there is only one solution, in 0<x<2pi.

@Deepak

It has two solutions as cos(x) = -1/2 has two solutions.

@||Mr.Insomaniac|| and siDDhESwaR MuKhErjEE

You say that cos(x) = -1/2 has two solutions.

But just for a minute check the domain of x for which i have taken the second case. x is in 2nd and 4th quadrant. cos(x) is positive in 4th quadrant. so, there is only one solution i.e. 2pi/3 in 2nd quadrant.

The second solution you are talking about is x = 4pi/3 which lies in third quadrant. So, it is not the solution.

Anyhow i have made the graph there you can clearly see thta there's only one solution.

Hope you agree with me this time.................

 YA GOT THE POINT