Starting from the origin a body oscillates simple harmonically with period of 2 s. After what time will its kinetic energy be 75% of the total energy

The kinetic energy of a body undergoing S.H.M. is given by   K.E. =12mω2A2cos2⁡ωt  or   (K.E.) max=12mω2A2According to given problem K.E.=(75/100)(K.E.)max∴ 12mω2A2cos2⁡ωt=7510012mω2A2  or  cos⁡ωt=±3/2  or  Ï‰t=π6  or  2πT×t=π6  or  t=T12=16s