solving an equation of the type: $t \sin (2t)=2$ where $0Need to solve:
How many solutions are there to the equation,
 $t\sin (2t)=2$ where  $0<t<3 \pi$
I am currently studying calc 3 and came across this and realized i dont have a clue as to how to get started on it.
 A: I'll be informal here:
Sketch the graph of $y=\sin(2t)$ over $(0,3\pi)$. The graph is composed of three $\sin$ waves with local maximums  at $x=\pi/4$, $x=5\pi/4$, and $x=9\pi/4$ and zeroes at $x=0$, $x=\pi$, $x=2\pi$, and $x=3\pi$.  
Now the graph of $y=t\sin (2t)$ is a "vertical scaling" of the graph of $y=\sin(2t)$. It has the same zeroes as before, and looks like three repetitions of a $\sin$ wave, except the amplitude increases as $x$ increases.
For reference, here is the graph.
We are interested in the number of points where this graph intersects the line $y=2$.
The first "hump", over the interval, $[0,\pi/4]$ lies below the line $y=2$: the maximum value of $y=t\sin(2t)$ over  $[0,\pi/4]$ is no greater than $\pi/4<2$. So, there are no intersection points in this interval.
The "humps" below the $x$-axis do not intersect the line $y=2$.
So, no point looking there.
This leaves us with two humps above the $x$-axis to consider--one over the interval $[\pi, 3\pi/2]$ and the other over the interval $[2\pi, 5\pi/2]$. Calculating $t\sin(2t)$ at the values $t=5\pi/4$ and $9\pi/4$ tells us that the "top" of both of these humps lie above the line $y=2$. Continuity then tells us that each of these humps intersects the line $y=2$ in (at least) two places.
We could be naive here, trusting the pretty graph in the link above and our intuition, and say there are exactly four intersection points. But we really should verify that there are no others with some degree of rigor.
To see that each hump gives exactly two intersection points, examine the derivative of $f(t)=t\sin(2t)$. Let's consider the middle hump and search for points where an extreme value of $f$ can occur in $(\pi,3\pi/2)$.  Calculating $f'$ and setting it equal to zero leads to the equation
$$
\tan(2t)=-2t.
$$
In the interval $(\pi,3\pi/2)$, this has only one solution, that corresponds to the top of the hump. (One can see this by examining the graphs of $y=-2t$ and $y=\tan(2t)$.) So, there is only one local extreme value of $f$ over $(\pi,3\pi/2)$. From this, it follows that there are exactly two intersection points in  $(\pi,3\pi/2)$. 
I'll leave the last hump to you.
The final line: there are four intersection points.
A: As an alternate approach, you could rewrite the equation as $$\frac{1}{2}\sin{2t}=\frac{1}{t}$$ and then observe that since $\frac{1}{t}\le\frac{1}{2}$ for $t\ge2$,
the graph of $y=\frac{1}{t}$ will intersect the graph of $y=\frac{1}{2}\sin{2t}$
twice in each interval $[n\pi,(n+1)\pi]$ for $n\ge1$.
