$x \sin x=2$ why is my proof that there no solutions wrong?

Question

$\frac 12 x \sin x=1$ . Let's look at a right triangle with base $x$ and altitude $\sin x$ . Then our equation is for the area of this triangle. Let the sides of the triangle be $a=x$ , $b=\sqrt {x^2+sin^2 x}$ , and $c= \sin x$ . According to wikipedia, Heron's formula can be written as $$A=\large \frac { \sqrt {4a^2c^2-(a^2+b^2-c^2)^2}}{4}$$

Plugging in:

$4=\large \sqrt{4x^2 \sin^2 x-(x^2+x^2+\sin^2 x-\sin ^2 x)^2}$

$4=x^2 \sin^2 x -x^4$

$x^2(x^2- \sin^2 x)=-4$

$x^2$ will always be positive, and $\sin^2 x$ is never greater than $x^2$ , so this equation can have no real solutions. The original has solutions, so why is this wrong?

Answer 1

This particular wikipedia formula is wrong.

It should be correctly either $$A=\large \frac { \sqrt {4a^2b^2-(a^2+b^2-c^2)^2}}{4}$$ or $$A=\large \frac { \sqrt {4a^2c^2-(a^2-b^2+c^2)^2}}{4}\,.$$ Mind the symmetry..

Answer 2

There are solutions on each interval $\left[2k\pi,2k\pi+\frac\pi2\right]$ for positive integer $k$ by the intermediate value theorem because $x\sin(x)$ is $0$ on the left end and $2k\pi+\frac\pi2$ on the right.

Heron's Formula should be $$ A=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{4} $$ Does that cause the same problem?