# How to solve -2sin(t)+ cos(t) =0

• Oct 23rd 2010, 09:30 AM
tinyone
How 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 AM
Jhevon
Quote:

Originally Posted by tinyone
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!

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 AM
wonderboy1953
Just a word of note: t may have more than one value.
• Oct 23rd 2010, 10:14 AM
tinyone
t=arctan(1/2) = 0.463647609

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

(Thinking)
• Oct 23rd 2010, 10:21 AM
e^(i*pi)
This isn't a special angle so you won't get it in terms of pi
• Oct 23rd 2010, 10:29 AM
tinyone
wonderboy 1953, how can I find the other solution...

is it perhaps done this way: 2π - 0.463647609

?
• Oct 23rd 2010, 10:53 AM
Jhevon
Quote:

Originally Posted by tinyone
wonderboy 1953, how can I find the other solution...

is it perhaps done this way: 2π - 0.463647609

?

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.