I really like Chris's answer to this question, using the graph plotter makes life so much easier.

Here's the 'on paper' method that we had to use before these sophisticated plotters were available.

Let

$$7\cos12t+6\sin12t\equiv R\cos(12t-\alpha)=R\cos12t\cos\alpha+R\sin12t\sin\alpha,$$

then

$$R\cos\alpha=7 \text{ and } R\sin\alpha=6.$$

Squaring and adding, $$R^{2}=7^{2}+6^{2}=85, \text{ so }R=\sqrt{85}.$$

Dividing, $$\tan\alpha = 6/7, \text{ so }\alpha = 0.7086 \text{ rad}.$$

We have then $$7\cos12t+6\sin12t=\sqrt{85}\cos(12t-0.7086)>9.$$

The maximum for the cosine occurs when 12t - 0.7086 = 0, that is, when t = 0.0591 (4dp, but I kept greater accuracy for all decimal values during the calculation).

For equality with the 9, $$12t - 0.7086 = \cos^{-1}(9/\sqrt{85}),\quad t=0.0773.$$

It follows that the wave is greater than 9 for the time span 2(0.0773 - 0.0591) = 0.0364.

Since the period of the wave is 2pi/12, the proportion of the time that the wave is greater than 9 is

0.0364/(pi/6) = 0.0696, that is, about 7% (agreeing with Chris).

As an alternative, the form $$R\sin(12t+\beta)$$ could have been chosen rather than the cosine.