How to solve $\sin x\cos x-x=0$

Question

How can I solve the equation:

$$\sin x\cos x-x=0\text{ ?}$$

They are not the same kind

(trig and polynomial).

Thanks.

There are rarely closed form solutions in this sort of situation. But here we have $\sin 2x=2x$, with obvious solution $x=0$, and no other solution since the absolute value of $sin t$ is less than the absolute value of $t$ except at $t=0$. — André Nicolas, Sep 06 '15 at 13:39
May be, this could be of interest to you (I hope and wish) http://math.stackexchange.com/questions/871516/how-could-i-improve-this-approximation — Claude Leibovici, Sep 06 '15 at 13:42

score 6 · Accepted Answer · answered Sep 06 '15 at 13:41

$$ \sin x\cos x = x $$ $$ 2\sin x \cos x = 2x $$ $$ \sin(2x) = 2x $$ $$ \sin u = u $$

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.

score 2 · Answer 2 · answered Sep 06 '15 at 13:39

2

$0$ 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 $$|x-\sin(x) \cos(x)| \geq |x|-|\sin(x) \cos(x)| >0$$

answered Sep 06 '15 at 13:39

N. S.

132,525

Aditya Dev · Answer 3 · 2015-09-06T13:45:15.300

1

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.

edited Sep 06 '15 at 13:45

answered Sep 06 '15 at 13:28

Aditya Dev

4,774

But still I'll have difficulty because it's not the same type. – Dave Sep 06 '15 at 13:29
it is the same type, he used the additional formuals – Dr. Sonnhard Graubner Sep 06 '15 at 13:30
How it's the same? One is trig $sin2x$ and the other is polynomial $2x$ – Dave Sep 06 '15 at 13:31
yes it is and your original equation is this also – Dr. Sonnhard Graubner Sep 06 '15 at 13:32
So the answer didn't help me, because it's still the same kind. – Dave Sep 06 '15 at 13:33
what kind do you exspect? we can't solve this equation explicit for $x$ – Dr. Sonnhard Graubner Sep 06 '15 at 13:33
Why not? For example $x=0$ is a solution – Dave Sep 06 '15 at 13:34
yes the is the only solution it comes from intuition but not from solving the equation by isolating $x$ – Dr. Sonnhard Graubner Sep 06 '15 at 13:36
So, we can't prove that it's the only solution? – Dave Sep 06 '15 at 13:38
I have edited it. – Aditya Dev Sep 06 '15 at 13:40
@Dave : It's not hard to prove it's the only solution. See my answer below. I suspect that what Dr. Sonnhard Graubner means is that one can't solve it simply by using algebra. ${}\qquad{}$ – Michael Hardy Sep 06 '15 at 13:43

score 1 · Answer 4 · answered Sep 06 '15 at 13:38

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 $y=x$.

How to solve $\sin x\cos x-x=0$

4 Answers4