# Some problems with trigonometry

Actually, this is the original question:

A particle moves along the x-axis so that the distance traveled in time $t$ is given by $x=2t + \cos 3t$. Find the distance between the first two positions of rest.

The ans given is 0.37 unit.

-
$2t+\cos\,3t=0$ is a transcendental equation; there isn't a general method to solve equations like these explicitly. One could use numerical methods for generating approximate solutions, though. – J. M. Sep 21 '11 at 13:08
@J.M. We do not need to solve $x(t)=2t+\cos(3t)=0$ at all. We need to solve $v(t) = 0$, where $v(t)= x'(t)$ is the velocity. The roots are just arcsin of some number. – Srivatsan Sep 21 '11 at 13:38
@Sri: I was addressing revision 1 of the question. Now I see OP edited to a different question. Not cool. – J. M. Sep 21 '11 at 13:45
Not you, @Sri. You had no fault. It was the OP's bait-and-switch that had me bummed. – J. M. Sep 21 '11 at 13:49
@Sophia When you are editing a question, please do not edit it so substantially that it will invalidate a comment or answer. If such an edit is necessary, then mention somewhere (either in the question itself or in the comments) that you have modified the question. Otherwise different people will be seeing/solving different revisions of the question, leading to unnecessary confusion. – Srivatsan Sep 21 '11 at 13:55

Introduce a time scaling by means of $t:={\tau\over3}$. Then we have to study the function $$x(\tau):={2\over 3}\tau+\cos\tau$$ with derivative $$\dot x(\tau)={2\over3}-\sin\tau\ .$$ Beginning at $\tau=0$ the first time of rest is at $\tau_1=\arcsin{2\over3}$, and the second at $\tau_2=\pi-\arcsin{2\over3}$. In between $\dot x(\tau)$ is negative; therefore the distance $d$ traveled in the time interval $[\tau_1,\tau_2]$ is given by \eqalign{d =x(\tau_1)-x(\tau_2)&={2\over3}\bigl(2\arcsin{2\over3}-\pi\bigr)+\cos\bigl(\arcsin{2\over3}\bigr)-\cos\bigl(\pi-\arcsin{2\over3}\bigr)\cr &={4\over3}\arcsin{2\over3}-{2\pi\over3}+{2\over3}\sqrt{5}\cr &\doteq 0.369287\ \cr }
As pointed out by Srivatsan's comment, the points of rest, where $x'(t)=0$, can be found to be: $$2-3\sin(3t)=0 \; \; \to \; \; t=\cases{ &\frac13 \left(2\pi n +\sin^{-1}\left(\frac23\right)\right)\\ &\frac13 \left(2\pi n +\pi +\sin^{-1}\left(\frac23\right)\right) }$$
What about $t=(1/3)(\pi-\arcsin(2/3))$? Oh, I think you may just have a plus sign where you need a minus in the second case. – Gerry Myerson Jul 24 '12 at 9:27