How many distinct values of x satisfy the equation sin(x) = x/2, where x is in radians?

Hello!

There is one obvious solution, `x_1=0.` Also, because `|sin(x)|lt=1,` there are no solutions for `|x|gt2.` And one more consideration, because `sin(x)` and `x/2` are both odd functions, for each positive solution `x` there is one negative solution  `-x.` So we can look only at `x in (0,2].`

We know that `sin(x)` is strictly concave down on `(0,pi/2),` and `sin(0)=0,` `sin(pi/2)=1.` Therefore `sin(x)` is greater than the secant line through `(0,0)` and `(pi/2,1).` The equation...

Hello!

There is one obvious solution, `x_1=0.` Also, because `|sin(x)|lt=1,` there are no solutions for `|x|gt2.` And one more consideration, because `sin(x)` and `x/2` are both odd functions, for each positive solution `x` there is one negative solution  `-x.` So we can look only at `x in (0,2].`

We know that `sin(x)` is strictly concave down on `(0,pi/2),` and `sin(0)=0,` `sin(pi/2)=1.` Therefore `sin(x)` is greater than the secant line through `(0,0)` and `(pi/2,1).` The equation of this line is `y=x*2/pi,` so `sin(x)gtx*2/pi>x/2` for `x in (0,pi/2).`

Well, there are no solutions on `(0,pi/2).` At `x=pi/2` `sin(x)=1gtpi/4=x/2,` at `x=pi` `sin(x)=0ltpi/2=x/2,` so by the Intermediate Value Theorem there is at least one solution on `(pi/2,pi).` We can find this solution `x_2` only approximately, it is about 1.895.

But from `x=pi/2` to `x=pi`  `sin(x)` decreases and `x/2` increases, so there is only one solution on `(pi/2,pi).`

So there are 3 solutions: `x_1=0,` `x_2 approx 1.895` and `x_3=-x_2 approx -1.895.`