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I haven't done math in 4 years now and I'm getting into Game programming. So I can't even remember where to start with this? I would like to solve for t.

$x=16\left(\sin \left(t\right)\right)^3$

ArmenB
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    Multiply both sides by $1\over {16}$, then extract the cubic root on both sides. Finally apply the inverse of $\sin$ on both sides, that is, use $\arcsin$. – Git Gud Sep 18 '13 at 22:24
  • Is this correct? $t=\sin \left(\left(\frac{x}{16}\right)^{\frac{1}{3}}\right)^{-1}$ – ArmenB Sep 18 '13 at 22:29
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    Almost, the notation is usually $\sin^{-1}(\cdots)$ or $\arcsin(\cdots)$. – abiessu Sep 18 '13 at 22:30
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    What you wrote is wrong, but I feel you got the right idea. To denote the inverse of $\sin$, if you're not using $\arcsin$, you should use $\sin ^{-1}$ and that would give you: $\displaystyle \sin ^{-1}\left(\left(\dfrac{x}{16}\right)^{1/3}\right)$. – Git Gud Sep 18 '13 at 22:31
  • I want to create the heart function which I calculated like so : $y=13\cdot \cos \left(\arcsin \left(\left(\frac{x}{16}\right)^{\frac{1}{3}}\right)\right)\space -\space 5\cos \left(2\cdot \arcsin \left(\left(\frac{x}{16}\right)^{\frac{1}{3}}\right)\right)-2\cos \left(3\cdot \arcsin \left(\left(\frac{x}{16}\right)^{\frac{1}{3}}\right)\right)\space -\cos \left(4\cdot \arcsin \left(\left(\frac{x}{16}\right)^{\frac{1}{3}}\right)\right)\space $ but when I graph it using https://www.desmos.com/calculator it's only graphing half of it. – ArmenB Sep 18 '13 at 22:36
  • That's probably due to domains of the functions and such. can't help you right now, I'm off. Try asking a new question and link it to this one. – Git Gud Sep 18 '13 at 22:39
  • If your equation has one solution $t$, it has infinitely many. You need some more restriction(s) on $t$. – Stefan Smith Sep 19 '13 at 00:10

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Do some algebra: $\sin^{-1}({\frac{x}{16}})^{1/3}$

$x = 16 \sin^3(t)$

$\frac{x}{16} = \sin^3(t)$

$(\frac{x}{16})^{1/3} = \sin(t)$

$(\sin^{-1}(x/16))^{1/3} = t$

Ross Millikan
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Maksim
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