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

[aaj92]

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 = \frac{arctan(-3/5)}{10} 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 :/

 

[Delphi51]

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.

 

[SammyS]

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 = \frac{arctan(-3/5)}{10} 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, \tan(x)=A\,, is x=\arctan(A)+k\pi\,, where k is an integer.