Turning a graph

Using complex numbers make the problem very easy to solve. The graph of $\sin(x)$ is the collection of points/numbers of the form $x+i\cdot\sin(x),\ \ x\in \mathbb{R}$ It is expressed in its polar form below as it is more useful for the solution: $r \cdot (\cos(\theta) + i \cdot \sin(\theta)),\ \ \ \ r = \sqrt{x^{2} + \sin^{2}(x)}, \ \ \theta = tan^{-1}\left(\frac{\sin(x)}{x}\right)$ Here, $r$ is called modulus and $\theta$ is called the argument of the complex number. When two complex numbers are multiplied, the arguments add and moduli multiply. Addition of angles indicate rotation. If a complex number of the form $\ \cos(\frac{\pi}{4}) + i \cdot \sin(\frac{\pi}{4})$ is multiplied with the set of numbers that make the graph of $\sin(x)$ , the whole graph will be rotated by $45^{\circ}$ (as the argument of the multiplied number is $\frac{\pi}{4}$ radians or $45^{\circ}$ and as multiplication of complex numbers cause angles to add) without any stretching as the modulus of the multiplied number is 1 (as $\sqrt{\cos^{2}(\theta) + \sin^{2}(\theta)} = 1$ ). Everything is shown below step by step: $\ \cos\left(\frac{\pi}{4}\right) + i \cdot \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot i$ $\left(x+i\cdot\sin(x)\right)\cdot \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot i \right) = \frac{x - \sin(x)}{\sqrt{2}} + \frac{x + \sin(x)}{\sqrt{2}} \cdot i$ Thus the parametric equation of $F$ is: $\left( \frac{t - \sin(t)}{\sqrt{2}}, \frac{t + \sin(t)}{\sqrt{2}} \right)$

Now, suppose, $x = \frac{k - \sin(k)}{\sqrt{2}}$ . Then, $F(x) = \frac{k + \sin(k)}{\sqrt{2}}$

So, $x \cdot F(x) = \left( \frac{k - \sin(k)}{\sqrt{2}} \right) \cdot \left( \frac{k + \sin(k)}{\sqrt{2}} \right) = \frac{k^{2}}{2} - \frac{\sin^{2}(k)}{2}$ Here, $x \cdot F(x) = \frac{\pi^{2}}{18} - \frac{3}{8} = \frac{\pi^2}{3^2 \cdot 2} - \frac{\sin^2( \frac{\pi}{3})}{2}$ So, $k = \frac{\pi}{3}$ . Solving for $x$ , we can get: $x = \frac{\pi}{3\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}} = 0.128108054...$

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