## Solving sin(t^2)-(t^2)=0

1. The problem statement, all variables and given/known data

Solve
sin(t2) - t2 =0. for t

2. Relevant equations

None, besides various trig identities. This was actually a dynamics problem where I had to solve for time, and this is simply the equation you get after summing up all the forces.

3. The attempt at a solution

Now obviously 0 is a solution, and when I plug it into wolfram alpha I get t=1.04, which I know is the correct answer because my professor told us that in class. However, when I plug it into my TI-89, the only solution that comes out is 0, and when I graph it, it doesn't show the 1.04 solution either.

So, my question is, is there any way to solve this problem without a numerical solver? And if not, is there any way to plug it into my TI 89 in order to get the correct answer?
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor

Recognitions:
Homework Help
 Quote by wutang42 1. The problem statement, all variables and given/known data Solve sin(t2) - t2 =0. for t 2. Relevant equations None, besides various trig identities. This was actually a dynamics problem where I had to solve for time, and this is simply the equation you get after summing up all the forces. 3. The attempt at a solution Now obviously 0 is a solution, and when I plug it into wolfram alpha I get t=1.04, which I know is the correct answer because my professor told us that in class. However, when I plug it into my TI-89, the only solution that comes out is 0, and when I graph it, it doesn't show the 1.04 solution either. So, my question is, is there any way to solve this problem without a numerical solver? And if not, is there any way to plug it into my TI 89 in order to get the correct answer?
t=1.04 can't possibly be a solution because

$$\sin(t^2)-t^2 < 0$$

for that value of t. How can you tell? Because the max sin can be is 1, and 1.042>1.
 for small radian values sin(x) ≈ x hence and that's probably the best you can do.

## Solving sin(t^2)-(t^2)=0

Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t2) - t2 =0

But yeah, I realize that it still doesn't make much sense, but this is the answer and the graph that wolfram alpha spits out, and I'm just trying to figure out how to replicate that either on paper or on my calculator
Attached Thumbnails

Recognitions:
Gold Member
Homework Help
 Quote by wutang42 Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t2) - t2 =0 But yeah, I realize that it still doesn't make much sense, but this is the answer and the graph that wolfram alpha spits out, and I'm just trying to figure out how to replicate that either on paper or on my calculator
Why doesn't it make much sense? Your graph clearly shows zeroes near ##\pm 1##. You could work it by hand with your calculator using Newton's method with a starting value of ##x=1##. Make sure your calculator is in radian mode.

 Quote by LCKurtz Why doesn't it make much sense? Your graph clearly shows zeroes near ##\pm 1##. You could work it by hand with your calculator using Newton's method with a starting value of ##x=1##.Make sure your calculator is in radian mode.
Aaaaand there's my problem. I was doing it in degrees the whole time on my calculator, and couldn't figure out why my graph looked nothing like the wolfram graph. Derp. Thanks!!

Recognitions:
Homework Help
 Quote by wutang42 Aaaaand there's my problem. I was doing it in degrees the whole time on my calculator, and couldn't figure out why my graph looked nothing like the wolfram graph. Derp. Thanks!!
Common mistake I've done it enough times myself that it's now on my checklist of things that could have gone wrong.