From your earlier questions, we found
[tex]2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)[/tex]
so the wave has amplitude √29. The weight's maximum negative position from equilibrium is then -√29, so you are solving for t in the given interval for which
[tex]\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac{\sqrt{29}}2[/tex]
Divide both sides by √29:
[tex]\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac12[/tex]
Take the inverse sine of both sides, noting that we get two possible solution sets because we have
[tex]\sin\left(\dfrac{7\pi}6\right)=\sin\left(\dfrac{11\pi}6\right)=-\dfrac12[/tex]
and the sine wave has period 2π, so [tex]\sin x=\sin(x+2\pi)=\sin(x+4\pi)=\cdots[/tex].
[tex]\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{7\pi}6+2n\pi[/tex]
OR
[tex]\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{11\pi}6+2n\pi[/tex]
where n is any integer.
Now solve for t :
[tex]t=\dfrac{\frac{7\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}[/tex]
OR
[tex]t=\dfrac{\frac{11\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}[/tex]
We get solutions between 0 and 0.5 when n = 0 of t ≈ 0.196946 and t ≈ 0.363613.
