A particle is executing SHM along a straight line and its position at time t is given by x=(4cos⁡2t+3sin⁡2t) cm. Then the particle arrives at its extreme position after a time of

x=4cos⁡2t+3sin⁡2t=42+32.cos(2t−tan⁡−1(34))∴V=dxdt=−5×2sin⁡(2t−tan⁡−134)cmsWhen the particle arrives at extreme position, its velocity becomes zero. ∴sin(2t−tan−134)=0⇒2t−tan−134=0⇒t=12tan−134  sec