ok first rewrite this as a single trig function with a phase offset.

$20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$

$A=\sqrt{20^2+4^2}=\sqrt{416}$

$\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$

Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for

$5t+\phi=\dfrac{2k+1}{2}\pi$

$t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$

you can probably take it from here.