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 $t$?
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 $t$?
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).
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?