A particle is executing SHM along a straight line and its position at time t is given by $x=(4cos2t+3sin2t)$ cm. Then the particle arrives at its extreme position after a time of 

$x=4\cos 2t+3\sin 2t=\sqrt{4^2+3^2}.\cos (2t-{\tan }^{-1}(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\left)\right)$

$\therefore V=\frac{dx}{dt}=-5\times 2\sin (2t-{\tan }^{-1}\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.)\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

When the particle arrives at extreme position, its velocity becomes zero. 

$\therefore \sin (2t-{\tan }^{-1}\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.)=0\Rightarrow 2t-{\tan }^{-1}\frac{3}{4}=0\Rightarrow t=\frac{1}{2}{\tan }^{-1}\frac{3}{4} \sec$