Solve for t: Forming a Series 3/(2√3)=cos(3t+4.189)

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The discussion centers on solving the equation 3/(2√3) = cos(3t + 4.189), which simplifies to 0.866 = cos(3t + 4.189). The solution involves recognizing that cos^(-1)(0.866) yields two angles: 30 degrees and 330 degrees. The correct approach is to express these angles in radians, leading to the equation 3t + 4.189 = 2nπ ± π/6, where n is any integer. This formulation allows for the derivation of a series of solutions for t.

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Homework Statement



3/(2√3)=cos(3t+4.189)

and therefore,

0.866=cos(3t+4.189)

I need to solve for t.

The Attempt at a Solution



I know that cos^(-1)θ has 2 values for every 2 pi, that is at 30 and 330 degrees.

But there is no limit given, so its probably from 0 to infinite,thus the question wants me to form a new series whereby by substituting n=0,1,2,3 and etc, I will be able to find t.

I don't know how to form this new formula/series as I was not thought this chap in calculus 2 (my lecturer was way behiind schedule, hence she skipped it)

Hope you guys can help me.

Thanks a lot.

 
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Instead of writing 30 and 330, think about 30 and -30 and angles coterminal with those. Using radians like you should in a problem like this you could express those angles as

[tex]3t + 4.189 = 2n\pi \pm \frac \pi 6,\ n = -\infty..\infty[/tex]

and solve for t. Is that 4.189 what you were really given or just a decimal approximation to it?
 

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