- 31 Mar '08 04:51 / 2 editsIt'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