thanks in advance
Hi !
Note that $\displaystyle \sin(4t)+\sqrt{3}\cos(4t)=2\left( \frac{1}{2}\sin(4t)+\frac{\sqrt{3}}{2}\cos(4t)\rig ht)
=2\left( \cos\frac{\pi}{3}\sin(4t)+\sin\frac{\pi}{3}\cos(4t )\right)$. As $\displaystyle \sin a \cos b+\sin b \cos a =\sin(a+b)$, we get that $\displaystyle \sin(4t)+\sqrt{3}\cos(4t)=2\sin\left( 4t+\frac{\pi}{3}\right)$ hence $\displaystyle x(t)=3+2\sin\left( 4t+\frac{\pi}{3}\right)$. Can you use this to answer the two questions?
Since the centre of motion is x = 3, you want the times at which x = 3:
$\displaystyle 3 = 3 + \sin (4t) + \sqrt{3} \cos (4t)$
$\displaystyle \Rightarrow 0 = \sin (4t) + \sqrt{3} \cos (4t)$
$\displaystyle \Rightarrow - \sin (4t) = \sqrt{3} \cos (4t)$
$\displaystyle \Rightarrow - \tan (4t) = \sqrt{3}$
$\displaystyle \Rightarrow \tan (4t) = - \sqrt{3}$.
And you know how to solve this trig equation ....?