# Solving for $x$ in $x(t) = \frac{-2}{3}\cos(10t) + \frac{1}{2}\sin(10t)$

A physics problem is asking me a to find when a weight on a spring crosses the equilibrium point.

The equation of motion given is $$x(t) = \frac{-2}{3}\cos(10t) + \frac{1}{2}\sin(10t)$$

Basically, I need to solve for $t$ when $x(t) = 0$. How do I solve for $t$ in such an equation?

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Hint Divide by $\cos(10t)$, as long as $\cos(10t)\neq 0$. –  Daryl Nov 12 '12 at 4:39
if you have $a \cos(\omega t)+b \sin(\omega t)=0$ then you take one term to the right hand side and divide both sides by say $\cos$ to get an equation in terms of tangent. Then you solve that equation. –  Maesumi Nov 12 '12 at 4:41

So, $\frac 2 3 \cos 10t=\frac 12 \sin 10t$

So, $$\frac{\cos 10t}{3} =\frac{\sin 10t}4$$

So, $\tan 10t=\frac 4 3\implies 10t=n\pi+\arctan \frac 4 3$ where $n$ is any integer.

So,$t=\frac{n\pi+\arctan \frac 4 3}{10}$

For $n=0,t=\frac{\arctan \frac 4 3}{10}$

Also, $$\frac{\cos 10t}{3} =\frac{\sin 10t}4=\pm\frac{\sqrt{\cos^2 10t+\sin^2 10t}}{\sqrt{3^2+4^2}}=\pm\frac 1 5$$

$\implies \cos 10t=\pm \frac 3 5,\sin 10t=\pm \frac 4 5$

$\cos 10t\cdot \sin 10t=\cos^210t\tan 10t=\cos^210t\cdot\frac 4 3>0$

So, the sign of $\cos 10t, \sin 10t$ will be same.

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So what's the first value of $t$ where $x(t)$ is zero? –  Imray Nov 12 '12 at 4:52
@Imray, please find the edited answer. –  lab bhattacharjee Nov 12 '12 at 4:55
@Imray, $x(t)=0\implies \frac{-2}{3}\cos(10t) + \frac{1}{2}\sin(10t)\implies \frac{2}{3}\cos(10t)=\frac{1}{2}\sin(10t)$, then divide either sides by $2$ –  lab bhattacharjee Nov 12 '12 at 5:04
Put $10t=\arctan \frac 4 3\implies \cos 10t=\pm \frac 3 5,\sin 10t=\pm \frac 4 5$ –  lab bhattacharjee Nov 12 '12 at 5:24
@Imray, sorry for the confusion.Please find the rectified answer. –  lab bhattacharjee Nov 12 '12 at 5:48