# $\sin x + x\cos x=0$ solution?

Any idea of solving this equation?

$$\sin x + x\cos x=0$$

I have also tried by setting a function $g(x)=\sin x+x\cos x$ and searching for solutions using the derivative but my atempts w

• Maybe the intermediate value theorem might help establish that there are indeed roots? – Kari Jul 14 '15 at 18:57
• $x = 0$ is one solution! In general we have $\tan x = -x$. This is a famous transcendental equation that has no closed form solution in general. – Simon S Jul 14 '15 at 18:57
• do you need only one solotion? – 3SAT Jul 14 '15 at 19:04
• Hint: $\sin x=\cos(90-x)$ and $\cos x=sin(90-x)$. – Chuks Jul 14 '15 at 19:17
• @SimonS why is it famous? – hjhjhj57 Jul 14 '15 at 19:19

## 3 Answers

First, we have the trivial solution: $$g(0) = \sin 0 + 0\cdot \cos 0 = 0$$

Consider $x = n\pi$ for large $n\in\mathbb{N}$, $$g(n\pi) = \sin(n\pi) + (n\pi)\cdot \cos(n\pi) = n\pi\cdot (-1)^{n}$$ We have infinitely many points where $g(x)$ is positive, and infinitely many points where $g(x)$ is negative. This coupled with the fact that the function is continuous and oscillating between positive and negative means that there are infinite solutions to your question.

A graph of $\frac{\sin x }{x}- \cos x$ shows a root at x = 0. Other roots asymptotically approach odd multiples of $\pi/2, ( x\rightarrow \infty)$. These can be found by numerical methods such as Newton-Raphson.

This kind of equations which mix polynomial and trigonometric functions do not show explicit solution (take into account that this is already the case for $x=\cos(x)$ and numerical methods should be used.

Narasimham explained that, beside the trivial solution $x=0$, the roots are closer and closer to odd multiples of $\frac \pi 2$ and they are closer and closer to this bound (just consider the positive roots since if $x$ is a solution of the equation, $-x$ is another one). This is then a good starting value $x_0$ for Newton method. The first iterates would then be $$x_0=\frac {(2k+1)\pi} 2$$ $$x_1=(2k+1)\frac \pi 2+\frac{2}{\pi (2 k+1)}$$ what you also could obtain using Taylor expansion at first order. You can, for sure, continue iterating to the required accuracy using $$x_{n+1}=x_n-\frac{\sin (x_n)+x_n \cos (x_n)}{2 \cos (x_n)-x_n \sin (x_n)}$$ Just for illustration purposes, the approximation $x_1$ gives for the first roots$$\{2.20742,4.92460,7.98131,11.0865,14.2079,17.3366,20.4693,23.6044\}$$ while the solutions would be $$\{ 2.02876, 4.91318, 7.97867, 11.0855,14.2074, 17.3364, 20.4692, 23.6043\}$$

Edit

Just for your curiosity, we could do better if, instead of using Taylor, we build the simplest Pade approximant of the function (don't worry : you will learn about them). In such a case, the solution would be approximated by $$x=(2k+1)\frac \pi 2+\frac{2 (2 k+1)\pi }{(2 k+1)^2 \pi^2 +6}$$ for which the numerical values are given below $$\{1.96672,4.91117,7.97828,11.0854,14.2074,17.3363,20.4691,23.6043\}$$