4
$\begingroup$

I'm trying to solve the following equation for $t$ in the first cycle

$0.8=-1.2\sin(2t)+0.8\cos(t)$

I've got it down to

$0.8=[\cos(t)](0.8-2.4\sin(t))$

Is there any algebraic way to continue this equation to solve for $t$?

  • 1
    Let's try this again: write $\sin(t) = \pm \sqrt{1 - \cos^2(t)}$. You need to be a bit careful about which quadrant contains $t$ (in order to determine the sign, that is).2011-04-21

2 Answers 2

1

Using DJC's suggestion, write $\sin(2t) = 2 \sin t \cos t$ and substitute $\sin t = \pm \sqrt(1-\cos^2 t)$, followed by $x = \cos t$ to get $2 (1-x) = 3 \cdot 2 x \sqrt(1-x^2)$ (can you see why we used the negative sign?) Squaring both sides, we get $1+x^2-2x = 9x^2(1-x^2)$, from which we can numerically find $x = 0.256431$ and, by inspection, $x=1$. The latter corresponds to $t=0$, the former to $t=\pi + \arccos 0.256431 = 4.453$ (shifted quadrant to get the right signs).

0

HINT: $\sin\left(2t\right)=2\sin\left(t\right)\cos\left(t\right)$

It seems you already noted this, you can further simplify to:

$0.8=0.8\cos\left(t\right)\left[-3\sin\left(t\right)+1\right]$

Do you see what to do from here?

  • 0
    @Paul Belardi - to get all the solutions one can square both sides and use the Pythagorean identity $\sin^{2}\left(x\right)+\cos^{2}\left(x\right)=1$ to turn this equation into a polynomial in $y=\sin\left(x\right)$. Unfortunately, you can factor a linear term out but you then end up with a cubic. You can actually solve the cubic equation in general via Cardano's formula. The trouble is that we have our solution in terms of $\sin(x)$ and $\sin(x)=k$ can in general only be solved numerically unless we get lucky and get a value we know. But that may be more than you wanted to know :)2011-04-22