Đáp án: $t=\arctan(\dfrac34)$ 

Giải thích các bước giải:

Ta có:

$OM^2=(3+5\sin t)^2+(4+5\cos t)^2$

$\to OM^2=30\sin \left(t\right)+40\cos \left(t\right)+25\sin ^2\left(t\right)+25\cos ^2\left(t\right)+25$

$\to OM^2=30\sin \left(t\right)+40\cos \left(t\right)+25(\sin ^2\left(t\right)+\cos ^2\left(t\right))+25$

$\to OM^2=30\sin \left(t\right)+40\cos \left(t\right)+25\cdot 1+25$

$\to OM^2=30\sin \left(t\right)+40\cos \left(t\right)+50$

$\to OM^2-50=30\sin \left(t\right)+40\cos \left(t\right)$

$\to (OM^2-50)^2=(30\sin \left(t\right)+40\cos \left(t\right))^2$

$\to (OM^2-50)^2\le (30^2+40^2)\cdot (\sin^2t+\cos^2t)$

$\to (OM^2-50)^2\le 2500$

$\to -50\le OM^2-50\le 50$

$\to 0\le OM^2\le 100$

$\to  OM\le 10$

$\to$Dấu = xảy ra khi $OM=10\to \dfrac{\sin t}{\cos t}=\dfrac{30}{40}$

$\to \tan t=\dfrac34$

$\to  t=\arctan(\dfrac34)$