Hi(Itwasntme), have to rearrange this equation for t, but keep getting it wrong. Can anyone help? Thanks and my gratitude

21=10 sin 10(t-9) + 11

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- Jun 21st 2010, 06:41 PMAlex1313Rearranging
Hi(Itwasntme), have to rearrange this equation for t, but keep getting it wrong. Can anyone help? Thanks and my gratitude

21=10 sin 10(t-9) + 11 - Jun 21st 2010, 06:48 PMundefined
- Jun 21st 2010, 06:56 PMAlex1313
first one. It should be 18. I just want to know how to solve for t with a trig function

- Jun 21st 2010, 07:36 PMundefined
I don't know why it would be 18.

$\displaystyle 21=10 \sin (10(t-9)) + 11$

$\displaystyle 10=10 \sin (10(t-9)) $

$\displaystyle 1= \sin (10(t-9)) $

$\displaystyle 10(t-9) = \frac{\pi}{2}+k\cdot2\pi, k\in\mathbb{Z}$

$\displaystyle t=9+\frac{\pi}{20}+k\cdot\frac{\pi}{5}, k\in\mathbb{Z}$ - Jun 21st 2010, 07:48 PMAlex1313
thanks anyways

- Jun 21st 2010, 07:58 PMundefined
Oh right, degrees.

$\displaystyle 21=10 \sin (10(t-9)^\circ) + 11$

$\displaystyle 10=10 \sin (10(t-9)^\circ) $

$\displaystyle 1= \sin (10(t-9)^\circ) $

$\displaystyle 10(t-9)^\circ = 90^\circ+k\cdot360^\circ, k\in\mathbb{Z}$

$\displaystyle t=18+k\cdot36, k\in\mathbb{Z}$

If you really wanted to you could write

$\displaystyle 10(t-9)^\circ = 90^\circ+k\cdot360^\circ, k\in\mathbb{Z}$

as

$\displaystyle 10(t-9)^\circ = \sin^{-1}(1)+k\cdot360^\circ, k\in\mathbb{Z}$