# Trig word problem - solving a trig equation.

• Mar 13th 2011, 06:17 PM
Sleight
Trig word problem - solving a trig equation.
I came across a word problem that I can't seem to puzzle through. I know the answer to it, but I can not seem to find out how to get to it.

"A block is set in motion hanging from a spring and oscillates about it's resting position x=0, according to the function $x=\sqrt{3}sin2t+cos2t$. For what values of t is the block at it's resting position x=0."

This is how I've worked it so far.

$0=\sqrt{3}sin2t+cos2t$

$-cos2t=\sqrt{3}sin2t$

$-\frac{cos2t}{sin2t}=\sqrt{3}$

$-cot2t=\sqrt{3}$

$tan2t=\frac{1}{\sqrt{3}}$

$2t=tan^-^1(\frac{1}{\sqrt{3}})$

$2t = 30\deg$

$t = 15\deg$

This answer, unfortunately, is not correct. I know that the correct answer is 75 degrees, but I have no clue how to get to that answer.

Any help would be appreciated. I've been puzzling over this question for a while now.
• Mar 13th 2011, 06:24 PM
Prove It
It's because if $\displaystyle -\cot{2t} = \sqrt{3}$ then $\displaystyle \tan{2t} = -\frac{1}{\sqrt{3}}$...
• Mar 13th 2011, 07:17 PM
Sleight
Quote:

Originally Posted by Prove It
It's because if $\displaystyle -\cot{2t} = \sqrt{3}$ then $\displaystyle \tan{2t} = -\frac{1}{\sqrt{3}}$...

Your completely correct. Not sure why I forgot about the negative sign, as I had it on the paper I was typing from. However, it still turns out as -15 deg, which doesn't work.
• Mar 13th 2011, 08:26 PM
Prove It
There will be a solution for $\displaystyle 2t$ in the second quadrant and the fourth quadrant.
• Mar 14th 2011, 05:46 AM
Sleight
Oh, now I'm even more confused. The solution that the book gives is either $75 \deg$, or $\frac{5\pi}{12}$, which are in quadrant one.
• Mar 14th 2011, 06:20 AM
Prove It
Because when you have the values of $\displaystyle 2t$, you need to divide them by $\displaystyle 2$ to get the values of $\displaystyle t$...
• Mar 14th 2011, 07:07 AM
Sleight
Ah, I understand now. I think I had some faulty notions at play. I was approaching the negative degree as when the spring was below the zero, but I guess the motion is always positive around the unit circle, hence the reason I had to get rid of the negative? Still having a hard time wrapping my head around the underlying concept, but it should click in soon.

Once again, Prove It, you have helped immensely. Thank you.
Now when I get home I am going to post a problem that has sin as an exponent, and I haven't got a clue. Not sure why this chapter in the book is so difficult for me.