Difficulty computing second derivative value in SHM problem

JDC123
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Homework Statement



The displacement of a machine is given by the simple harmonic motion as x(t) = 5cos(30t)+4sin(30t). Find the amplitude of motion, and the amplitude of the velocity.

Homework Equations



x''(t) = -4500cos(30t)-3600sin(30t)

The Attempt at a Solution


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I should note that I'm not having any difficulty with the problem conceptually, as it is very simple. However, for some reason unknown to me, I'm getting the wrong sign on a t-value when I set x''(t) to zero. I've checked and rechecked my work numerous times, and I'm not sure what's going on. This exact same procedure worked with the first derivative, so I don't get why it's failing me with the second. I've graphed x''(t) on Desmos, and the first positive t-value for x''(t)=0 is 0.0749. My answer is equal to 0.02986. Strangely, according to the graph, the first negative t-value to satisfy x''(t)=0 is -0.02986. Does anyone know why I'm getting the wrong sign? It's driving me absolutely insane. Please see the attached photo to view my attempt at a solution.

It is admittedly quite late here, and while there must be some pernicious error in my algebra, I can't find it.Thank you!
 

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It is unclear what you mean when you say that you want to find the x-values for which x'' is zero. This should not happen for negative or for positive x. Try writing x" as a function of x instead of t.
 
My apologies! I'm after t-values for x''(t)=0, not x-values. I have edited my original post to correct this mistake.
 
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When you square an equation you must check that the solutions satisfy the original equation. Later you are missing one of the solutions when taking the square root and the solution you are left with happens to be the false root you introduced by squaring. I would quote the specific point in your work but this is impossible since you just posted an image. Generally, note that only posting photos of your work breaks the PF guidelines for homework and that such posts may be deleted.
 
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