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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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    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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    @Imray, sorry for the confusion.Please find the rectified answer.2012-11-12