How many times are the hands of a clock at 33 degrees? How many times are the hands of a clock positioned such that there is an angle of $33°$ between them in a day ?

My trying:
The minute hand moves 360 degrees in 60 minutes. This means that the angle of the minute hand is given by 6t, where t is number of minutes past midnight.
The hour hand moves 30 degrees in 60 minutes. This means that the angle of the hours hand is given by 0.5t.
The hands start together at midnight. The first time they make a 33 degree angle is when the minute hand has moved 33 degrees further than the hour hand, so this is given by the equation:
$ 6t = 0.5t + 33 \Rightarrow 5.5t = 33 \Rightarrow t = 6 minutes$ . In other words about 6 minutes past midnight.
I have found that 22 times the hour and minutes hand are in 33 degree angle . How can I find this number ? How can I find the 22 times ? 
 A: This problem does not require any complex math to solve: during one day, the hour hand goes around twice, and the minute hand 24 times, in the same direction. As such, the two hands will be in exactly the same position 22 times. That is, there will be a difference of $0$ degrees between them $22$ times. But, any other difference of degree, say $x$ degrees, will happen exactly that many times as well, because that difference will happen exactly once between two successive times when the difference is $0$. So, no matter what difference of degrees we are talking about, it will happen $22$ times a day.
A: So, after $t$ minutes, the difference of angles $$6t-\dfrac t2=\dfrac{11t}2$$
We need $$\dfrac{11t_r}2=360^\circ r\pm33^\circ$$ where $r$ is any integer
$$\iff t_r=\dfrac{2(360^\circ r\pm33)}{11}$$
So, $t_{n+1}-t_n=\dfrac{720}{11}$
So, in $24$ hours, each these two $+33^\circ,-33^\circ$ is occur $$\dfrac{24\cdot60}{\dfrac{720}{11}}=22$$ times
A: Let $M$ be the minute hand and $H$ the hour hand. Then we have that the angular speed of $M$ is $6°$ per minute and that of $H$ is $0.5°$ per minute. Thus $M$ moves $12$ times as fast as $H$.
Now when $M$ has turned $\theta$ degrees from midnight, $H$ has only covered $\theta/12$ degrees. We want $\theta$ so that $$\theta-\frac\theta{12}=33.$$ Hence $M$ is always $\theta=36°$ away from the last point where the constraint  $\theta-\theta/12=33$ was satisfied.
In one complete turn of $M$ therefore, this angular difference between $H$ and $M$ occurs $360°/36°=10$ times, so that -- since there are $24$ full turns of $M$ in a day -- the number of times the angle between $H$ and $M$ is $33°$ is $240.$
