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 

 $\text{ K.E. }=\frac{1}{2}{\mathrm{mω}}^2{\mathrm{A}}^2{\cos }^2⁡\mathrm{ωt}\text{ or }{\text{ (K.E.) }}_{max}=\frac{1}{2}{\mathrm{mω}}^2{\mathrm{A}}^2$
According to given problem
 $\mathrm{K}.\mathrm{E}.=(75/100)(\mathrm{K}.\mathrm{E}.{)}_{max}$
$$\begin{gathered} ∴\frac{1}{2}m{\mathrm{ω}}^2{A}^2{\cos }^2⁡\mathrm{ω}t=\left(\frac{75}{100}\right)\left(\frac{1}{2}m{\mathrm{ω}}^2{A}^2\right) \\ \text{ or }\cos ⁡\mathrm{ω}t=Â\pm \sqrt{3}/2 \\ \text{ or }\mathrm{ω}t=\frac{\mathrm{π}}{6}\text{ or }\frac{2\mathrm{π}}{T}×t=\frac{\mathrm{π}}{6}\text{ or }t=\frac{T}{12}=\frac{1}{6}\mathrm{s} \end{gathered}$$