Need help with finding positive time for x(t)= cos(wt) + sin(wt) problem

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Homework Statement



This is for a physics class but at this point in the problem it's basic math... which for some reason I can just not figure out. I need to find at what time x(t) = 0 and I keep getting the negative time... I want the first positive time.

Homework Equations



x(t) = 3cos(10t) + 5sin(10t) = 0


The Attempt at a Solution



i get that t = [itex]\frac{arctan(-3/5)}{10}[/itex] which equals -0.054 seconds. Since you can't really have negative time the answer should be 0.26 seconds which is shown if this equation is graphed. I know this is dumb but I cannot figure out how to get this answer. I'm pretty sure you just have to shift it a certain amount... I just don't know what that amount is :/
 
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If you write it as z = 3*cos(A) + 5*sin(A) = 0
and do the old transformation method for solving trigonometric equations: x = cos(A), y = sin(A) it becomes 3x + 5y = 0. This represents a line on the graph, and the intersections of the line with the unit circle are the solutions. Clearly there are two solutions for A, one positive and one negative.
 
aaj92 said:

Homework Statement



This is for a physics class but at this point in the problem it's basic math... which for some reason I can just not figure out. I need to find at what time x(t) = 0 and I keep getting the negative time... I want the first positive time.

Homework Equations



x(t) = 3cos(10t) + 5sin(10t) = 0


The Attempt at a Solution



i get that t = [itex]\frac{arctan(-3/5)}{10}[/itex] which equals -0.054 seconds. Since you can't really have negative time the answer should be 0.26 seconds which is shown if this equation is graphed. I know this is dumb but I cannot figure out how to get this answer. I'm pretty sure you just have to shift it a certain amount... I just don't know what that amount is :/

The general solution to the equation, [itex]\tan(x)=A\,,[/itex] is [itex]x=\arctan(A)+k\pi\,,[/itex] where k is an integer.