An equation involving trigonometric functions.

Question

$$\sin(x) - x\cos(x) - \frac{1}{2}x^2 = 0$$ I can verify that 0 is the solution of equation. But i don't have any method to show that is the only solution.

Answer 1

The first derivative of the LHS is

$$x(\sin x-1)$$ and it cancels for $x=0$ and $x=\dfrac\pi2+2k\pi$, giving the extrema.

For these values,

$$f(x)=0$$ and

$$f(x)=1-\frac12\left(\dfrac\pi2+2k\pi\right)^2,$$ which are all negative values.

Answer 2

HINT

Let indicate

$$f(x)=\sin x -x\cos x-\frac12 x^2 \implies f'(x)=x\sin x-x=x(\sin x-1)$$

Answer 3

As you said, $x=0$ is a solution of the equation. Let us calculate the derivative of the function $f(x)=\sin(x)-x\cos(x)-\frac{1}{2}x^2$:

$$f'(x)=x(\sin(x)-1)$$

The derivative tells us if a function increases or decreases. Since $|\sin(x)|\le 1$, it is clear that $\sin(x)-1$ is always negative. This means that $f'(x)$ is always positive if $x<0$ and negative if $x>0$. So the function is increasing until $x=0$ and then decreasing, meaning that $x=0$ is the global maximum. Since $f(0)=0$, we know that $0$ is the maximum of the function that is only reached once for $x=0$.