1. ## Linear Trigonomic Equation

Just a word problem thats bugging me:

The height, in metres, of a nail in a water wheel above the surface of the water, as a function of time, can be modelled by: h(t)=-4sin pi/4 (t-1) + 2.5 where t is time in seconds. During what periods of time is the nail below the water in the first 24 s that the wheel is roatating...

There is a similar proble, in the book which outlines how you would solve it, except that there are no brackets (where the t-1 is) so this is throwing me off, because I don't know how to get to the next step after that and everything i've tried has been wrong.

2. Originally Posted by abc10
Just a word problem thats bugging me:

The height, in metres, of a nail in a water wheel above the surface of the water, as a function of time, can be modelled by: h(t)=-4sin pi/4 (t-1) + 2.5 where t is time in seconds. During what periods of time is the nail below the water in the first 24 s that the wheel is roatating...

There is a similar proble, in the book which outlines how you would solve it, except that there are no brackets (where the t-1 is) so this is throwing me off, because I don't know how to get to the next step after that and everything i've tried has been wrong.

you were already provided with a most thorough solution ... what/where is your problem with it?

S.O.S. Mathematics CyberBoard :: View topic - Linear Trig Equation Word Problem

3. Wow! Haha, I didn't know that page existed, I didn't post that but obviously its sufficient. Thank you.

4. Why not say

$0=-4\sin \frac{\pi}{4} (t-1) + 2.5$

$-2.5=-4\sin \frac{\pi}{4} (t-1)$

$0.0625 = \sin \frac{\pi}{4} (t-1)$

$\sin^{-1}(0.0625) = \frac{\pi}{4} (t-1)$

$\frac{4}{\pi}\sin^{-1}(0.0625) = t-1$

$\frac{4}{\pi}\sin^{-1}(0.0625)+1 = t$

Is this the only soln? ;0)