# A trig question

joe shmo
Posers and Puzzles 31 Mar '08 03:51
1. joe shmo
Strange Egg
31 Mar '08 03:512 edits
How is the following equation solved

2 sin(2t) - sqrt(2)tan(2t) = 0

I come up with cost = sqrt(2)/4 & sint = sqrt(14)/4

I think I'm wrong?

If i am right , how do you derrive the t values for [0, 2pi]
2. 31 Mar '08 04:512 edits
It's pretty simple actually.

{x} = the square root of x

2 sin(2t) - {2}tan(2t) = 0 : Factor out sin(2t)
sin(2t)[2 - {2}sec(2t)] = 0

Two possibilities:

Possibility (1): 2 - {2}sec(2t) = 0 => {2}sec(2t) = 2 => sec(2t) = {2} => cos(2t) = {2}/2
=> 2t = pi/4, 7pi/4, 9pi/4, 15pi/4
t = pi/8, 7pi/8, 9pi/8, 15pi/8

Possibility (2): sin(2t) = 0 => 2t = 0, pi, 2pi, 3pi, 4pi
=> t = 0, pi/2, pi, 3pi/2, 2pi

So the solution set is:
t = 0, pi/8, pi/2, 7pi/8, pi, 9pi/8, 3pi/2, 15pi/8, 2pi
3. 31 Mar '08 05:30
as Jirakon used, but it may not be entirely obvious, the identity tan(2t) = sin(2t)/cos(2t) was used at the beginning.

then factoring the sin(2t) leads to 2 - sqrt(2)/cos(2t) or, more simply: 2 - sqrt(2)*sec(2t).

hope this helps