A question asks at what time of day is the rate of power consumption the greatest. t>=0 with noon coresponding t=0

The rate of consumption I derived, which is correct is $\displaystyle 450cos(\frac{pi*t}{12})+700$

I know I have to set the Rate equal to zero to get the maximum

$\displaystyle 450cos(\frac{pi*t}{12})+700=0$

$\displaystyle 450cos(\frac{pi*t}{12})=-700$

$\displaystyle cos(\frac{pi*t}{12})=\frac{-700}{450}$

$\displaystyle arccos(cos(\frac{pi*t}{12}))=arccos(\frac{-700}{450})$

$\displaystyle \frac{pi*t}{12}=arccos(\frac{-700}{450})$

I stopped right here because my calculator didnt like the Arccos(700/450). Can I get a point in the right direction?