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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 $\cos(10t)\neq 0$.2012-11-12
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    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.2012-11-12

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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?2012-11-12
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    @Imray, please find the edited answer.2012-11-12
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    So when $t$ is $\approx .0927$, the equation should be equal to zero? But when I plug it the value it's still pretty far from zero.2012-11-12
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    @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$2012-11-12
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    Why am I not getting zero when I set $t = arctan \frac{4}{3}$2012-11-12
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    Put $10t=\arctan \frac 4 3\implies \cos 10t=\pm \frac 3 5,\sin 10t=\pm \frac 4 5$2012-11-12
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    ok Thanks it works now!2012-11-12
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    It's not clear how you went from $\pm \frac{1}{5}$ to the next step... Where did the $tan(10t)$ come from?2012-11-12
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    @Imray, sorry for the confusion.Please find the rectified answer.2012-11-12