2sinx - x = 0?

Question

I forgot to add the 2 to the equation previously. So apart from the point (0,0). Is it possible to determine the other x-intercept for the curve y = 2sinx -x algebraically without using approximation techniques.

Answer 1

Yeah. You know that the regular graph of y=sin(x) has x-intercepts of 180, 360, 540 .. etc

Now since you know the period is not changing, for every 180 degrees, there will be an x-intercept. So just plug them in for your x-values.

Answer 2

Not in closed form. 2sinx-x has exactly 2 solutions other than (0,0), one at A, say, and another at -A, at each of these points.

However there is no "nice" way of writing A in terms of numbers and constants (like pi) that we know.

It is similar to asking if it is possible to determine the x-intercepts of y=sin(x). You would probably say, yes, the intercepts are just pi*integer, for any integer.

But what is pi? Pi is effectively defined so that 2*pi is the period of the sin function.

There is no way to write pi "algebreically" without using approximation techniques.
Similarly there is no way to write A "algebreically" without using approximation techniques.
(This is because pi and A are transcendental numbers)

Answer 3

No it is not possible
y=2sinx-x so if = 0 sinx=x/2 so-2<=x<=2
at x= -2
y(-2) =-2sin2+2>0
y(2) =2sin2-2 <0
y´=2cosx-1 =0 cos x= 1/2 and x= -pi/3 and pi/3
signy´------- (-pi/3++++++(pi/3) -------
y(-pi/3)=-sqrt3+pi/3 <0 so there is a root between -2 and -pi/3
f(pi/3)= sqrt3-pi/3 >0 so there is a root which is 0
There is another root >pi/3 as the function is an odd one