Isolating the term in 2nd order differential equation with complex roots.

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This is the initial value problem.

$5y''+ 2y'+ 7y = 0, \qquad y(0)=2, \quad y'(0)=1.$

Solution:

$ y = e^{-\frac{x}{5}} \cdot \left[2\cos\left(\frac{\sqrt{34}}{5} x\right)+ \frac{7}{\sqrt{34}}\sin\left(\frac{\sqrt{34}}{5}x\right)\right] $

Question: Which is the smallest T, for $|y(x)|\leq 0.1$ for all x > T.

I think I should isolate the x term in the the equation:

$ 0.1 = e^{-\frac{x}{5}} \cdot \left[2\cos\left(\frac{\sqrt{34}}{5} x\right)+ \frac{7}{\sqrt{34}}\sin\left(\frac{\sqrt{34}}{5}x\right)\right] $

But I don't know how to isolate x. Someone knows how to accomplish this?

Answer: T = 14.5115

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$$ |y| = e^{-\frac{x}{5}} \cdot \left |\bigg[2\cos\left(\frac{\sqrt{34}}{5} x\right)+ \frac{7}{\sqrt{34}}\sin\left(\frac{\sqrt{34}}{5}x\right)\right]\bigg|$$

$$ = e^{-\frac{x}{5}} \sqrt {4+49/34}$$

Now you want $$ e^{-\frac{x}{5}} \sqrt {4+49/34}<0.1$$

Can you solve it from here?