cot x/2=sin x interval[0,2piy]

Question

cot x/2=sin x interval[0,2piy]

Answer 1

 

You can easily solve this equation as follows:

 

First, notice that

 

cot (x/2) = cos(x/2) / sin(x/2)

 

and

 

sin(x) = 2*sin(x/2)*cos(x/2).

 

Hence, the equation that we want to solve is:

 

cos(x/2) = 2*cos(x/2)*[sin(x/2)]^2.

 

The equation is solved if either

 

A) cos(x/2) = 0, and, therefore, x/2 = pi/2, 3*pi/2, ....

 

But, as we are asked to solve in the interval [0, 2*pi) the only valid solution in this case is x = pi.

 

B) 1 = 2*[sin(x/2)]^2 and therefore sin(x/2) = sqrt(1/2) or sin(x/2) = -sqrt(1/2).

These has the possible solutions:

 

x/2 = pi/4, 3*pi/4, 5*pi/4, 7*pi/4,...

 

But, as we are asked to solve in the interval [0, 2*pi) the only valid solutions in this case are x= pi/2 and x = 3*pi/2.

 

Let me know if you have any further question.