How do I solve $\sin x+x\cos x=0\,$?

Question

How do I solve $\sin x+x\cos x=0\,$?

I've tried several different trigonometric identities and I'm aware it can be also written as $$\tan x=-x.$$ One of the answers is zero, but the other answers elude me. I've tried graphing $\tan x$ and $y=-x$ simultaneously as well, but I want to find the answer numerically.

The graph $y = \tan x$ and $y=-x$ intersect infinitely many time, one in every $(-\pi/2, \pi/2)+n\pi$. — Arctic Char, Dec 14 '21 at 20:00
Try defining $f(x)=\tan(x)+x$ and proving that it is strictly increasing using the derivative. — PQH, Dec 14 '21 at 20:03
I want to find the answer numerically --- See Numerical solution to x = tan (x) AND Derivation of asymptotic solution of $\tan(x) = x$. AND Solution of tanx = x? — Dave L. Renfro, Dec 14 '21 at 20:13
@PQH What would that show us, precisely? The strict monotonicity and continuity of $f$ wouldn't imply injectivity since $f$ isn't defined in every $\pi/2 + n\pi$ ... — Neckverse Herdman, Dec 14 '21 at 21:03
@Vardenmeersch it shows that the solution $x=0$ is unique at an interval. Then you can try the same at the other intervals. — PQH, Dec 15 '21 at 12:39
This is the transcendental equation, which does not have an analytical solution. You can solve it numerically using the Newton method. Define $f(x)=\text{tan}(x)+x$, set $x_0$ and solve $x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}$. Solve it for different $x_0$ (see graph of $f(x)$ to know where to start). — Vítězslav Štembera, Dec 19 '21 at 19:54

Claude Leibovici · Answer 1 · 2021-12-15T06:34:51.903

2

To avoid the discontinuities introduced by the tangent, keep the equation as $$\sin( x)+x\cos (x)=0$$ The roots will be closer and closer to $x_0^{(n)}=(2n+1)\frac \pi 2$. So, make series expansion around this point and then series reversion.

Have a look here for simple approximations.

edited Dec 15 '21 at 06:34

answered Dec 15 '21 at 05:37

Claude Leibovici

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How do I solve $\sin x+x\cos x=0\,$?

1 Answers1