I have: -2sin(t)+ cos(t) =0

How can I find out what t is?

I dont even know how to start, because I havent done this in over 3 years..

Help would be appreciated!

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- Oct 23rd 2010, 09:30 AMtinyoneHow to solve -2sin(t)+ cos(t) =0
I have: -2sin(t)+ cos(t) =0

How can I find out what t is?

I dont even know how to start, because I havent done this in over 3 years..

Help would be appreciated! - Oct 23rd 2010, 09:41 AMJhevon
first note that neither sin(t) nor cos(t) can be zero here (do you see why?)

since they are not zero, we can divide by either of them. hence,

$\displaystyle \displaystyle -2 \sin t + \cos t = 0$

$\displaystyle \displaystyle \Rightarrow 2 \sin t = \cos t$

$\displaystyle \displaystyle \Rightarrow 2 \tan t = 1$ ..............divided by the cos(t)

$\displaystyle \displaystyle \Rightarrow \tan t = \frac 12$

Now what? - Oct 23rd 2010, 10:00 AMwonderboy1953
Just a word of note: t may have more than one value.

- Oct 23rd 2010, 10:14 AMtinyone
t=arctan(1/2) = 0.463647609

But i need the answer to be in π (pi)...

(Thinking) - Oct 23rd 2010, 10:21 AMe^(i*pi)
This isn't a special angle so you won't get it in terms of pi

- Oct 23rd 2010, 10:29 AMtinyone
wonderboy 1953, how can I find the other solution...

is it perhaps done this way: 2π - 0.463647609

? - Oct 23rd 2010, 10:53 AMJhevon
don't approximate arctan(1/2), leave it as arctan(1/2)

there are infinitely many solutions, not just 2--not unless the problem itself tells you to restrict your answer.

Hint 1: tangent is periodic.

Hint 2: tangent takes on positive values in the first and third quadrants. arctan(1/2) is in the first quadrant. apply what you know of reference angles.