trigonometry equations 
Take this question:
"We follow the tips of the hands of an old fashioned analog clock (360 degrees is 12 hours) . We take the clock and put it into an axis system. The origin (0,0) of the axis system is the rotationpoint of the hands. The positive x-as goes through "3 hour" and the positive y-as through "12 hour" We calculate the time "t" in hours, starting from 0:00 hours. 
The equation for the tip of the big clockhand is: x=3sin2πt, y=3cost2πt 
The equation for the tip of the small clockhand is x=2sin(1/6)πt, y=2cos(1/6)πt  
On t=0 the two hands overlap eachother. Calculate the first point in time after t=0 when this occurs.
SOLUTION:
"This is true when cos(2πt)=cost(1/6πt) and sin(2πt)=sin(1/6πt) 
So, t = 12/11"
I simply don't know where to start...
 A: For the overlap thing, the next time there's an over lap is when the minute hand has already gone through a full hour (revolution).  Then
$$2 \pi (t+1) = \frac{\pi}{6} t \implies t + 1 = \frac{12}{11}$$
A: Let our time $t$ be $x$ hours and $y$ minutes where $0 \leq x < 12$ and $0 \leq y < 60$.
Let $\theta_1$ be the angle of the minute hand and $\theta_2$ be the angle of the hour hand where we measure angles clockwise beginning at 12 o'clock so that $0 \leq \theta_1, \theta_2 < 360^{\circ}$
It follows that $\theta_1 = 360 \times \frac{y}{60}$ and $\theta_2 = 360 \times \frac{x}{12} + 360\times \frac{y}{60\times 12} = 360(\frac{x}{12} + \frac{y}{60\times 12})$.
If the angle of the hour hand is the same as the minute hand, we have $\theta_1 = \theta_2$ and hence:
$$ \frac{y}{60} = \frac{x}{12} + \frac{y}{60\times 12}$$
We know that the hour and minute hands will have equal angles at some time shortly after 1 o'clock, so setting $x=1$ gives:
$$\frac{y}{60} - \frac{y}{60 \times 12} = \frac{1}{12} \iff \frac{12y-y}{60\times 12} = \frac{1}{12} $$
Solving for $y$:
$$11y = 60 \iff y =  60/11$$
So the two angles are equal $1$ hour and $60/11$ minutes after 12 o'clock, which is $1 + 1/11 = 12/11$ hours after 12 o'clock.
