How to solve $\sin x\cos x-x=0$ How can I solve the equation:
$$\sin x\cos x-x=0\text{ ?}$$
They are not the same kind
(trig and polynomial).
Thanks.
 A: $$
\sin x\cos x = x
$$
$$
2\sin x \cos x = 2x
$$
$$
\sin(2x) = 2x
$$
$$
\sin u = u
$$
Draw the graphs of $y=\sin u$ and $y=u$ superimposed on each other.  You see that $y=0$ when $u=0$ in both graphs.  That says $u=0$ is one solution.  The question is whether there are others.
Notice that the slope of $\sin u$ with respect to $u$ is $1$ when $u=0$, since $\sin'=\cos$.  Then notice that the slope of the sine function is smaller at all other points than it is at points where its slope is $1$.  That tells you the graph of the sine function must be below the graph of $y=u$ when $u>0$ and above the graph of $y=u$ when $u<0$.  That tells you whether there are any other solutions.
A: $0$ is the only solution. 
If $x \neq 0$ we know that $| \sin(x) | < |x|$ and $|\cos(x)| \leq 1$. This implies that
$$
|\sin(x)\cos(x)| < |x|$$
and hence 
$$|x-\sin(x) \cos(x)| \geq |x|-|\sin(x) \cos(x)| >0$$
A: Use the property 
$2\sin x\cos x=\sin (2x)$
Let $f(x)=sin2x - 2x$
$f'(x)=2\cos (2x)-2$ which is zero at $0,2\pi,...$
Also, $-1-2x\le f(x)\le 1-2x$ which means the function is contained between two lines.
Finally, $f'(x)\le 0$ which means the function is decreasing.
Also, $f(-\frac{\pi}{4})>0$ and $f(\frac{\pi}{4})<0$.This means, there has to be at least one point where it crosses the x-axis.
Since the curve decreases continuously, it intersects x-axis only once.
For solving such questions, try drawing the graph to find roots.
A: The equation $\sin u=u$ has only one solution: $u=0$. To see this, you may suppose $u>0$ because if $u<0$ is a solution, $-u$ is also a solution. You also may suppose 01$ and $\sin u\le 1.
Now, on the interval $[0,\frac\pi2]$, the sine function is strictly concave, hence its curve is under its tangent at any point in the interval, in particuler under its tangent at origin, which has equation $y=x$.
