Hi, 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
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)) $
$\displaystyle 10(t-9) = \frac{\pi}{2}+k\cdot2\pi, k\in\mathbb{Z}$
$\displaystyle t=9+\frac{\pi}{20}+k\cdot\frac{\pi}{5}, k\in\mathbb{Z}$
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
$\displaystyle 10(t-9)^\circ = \sin^{-1}(1)+k\cdot360^\circ, k\in\mathbb{Z}$