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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 $t$?

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You should factor your original equation and use the fact that $AB = 0 \Longrightarrow A = 0$ or $B = 0$. –  JavaMan Apr 21 '11 at 18:32
    
I don't see how to factor the original with 3 terms, one being sin(2t), one being cos(t) and one being a constant –  Paul Belardi Apr 21 '11 at 18:34
    
My mistake. I thought the original problem said $\sin(t^2)$, not $\sin(2t)$. –  JavaMan Apr 21 '11 at 18:38
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). –  JavaMan Apr 21 '11 at 18:44

2 Answers 2

up vote 1 down vote accepted

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).

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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?

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Note: the solution I have in mind does not exhaust all solutions, and the way I am thinking about this problem, it seems difficult to find exact values for all solutions. I can elaborate on this point if you are interested, but I felt it may have been a bit too distracting. –  WWright Apr 21 '11 at 18:58
    
The simplified form you have posted is identical to the simplified form I have posted had I factored 0.8 from the RS. Therefore I am still stuck at this point. –  Paul Belardi Apr 21 '11 at 19:18
    
@Paul Belardi - I was hoping this would make it clear to take t=0, but I guess its something you either see or don't see. Note integer multiples of $2\pi$ also solve this. –  WWright Apr 21 '11 at 20:28
    
You're right, it is clear now. I was hoping for a solution that would suffice in a more general case, solving for equations with sin(2t) and cos(t) where t=0 is not a solution. I take the fault here as I should have reworded my question to the more general case. –  Paul Belardi Apr 22 '11 at 1:10
    
@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 :) –  WWright Apr 22 '11 at 1:51

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