Question 27039
Hi,

There is probably a trig identity that solves this immediatly, but I'm useless at remembering them, I only know 3 or 4 myself. So I looked for a cleverer solution. And after staring for a few minutes, I found it.

Square your equation, and see what you get.

*[tex (\sin 2x - \cos 2x)^2 = 1^2]
*[tex \sin^2 2x + \cos^2 2x -2\sin 2x\cos 2x = 1]

Now using, *[tex \cos^2 + \sin^2 = 1], and subtracting one from both sides, we get the much nicer

*[tex -2\sin 2x\cos 2x=0]

This can be solved either by using the sine addition formula and saying that *[tex \sin 4x=0] or by saying that either *[tex \sin 2x=0] or *[tex \cos 2x=0]. Either way you should get the solutions as

*[tex x=\frac{\pi}{2}n,\frac{\pi}{4}(1+2n),n\in\{0,1,2,3}]

'Wait! Zero isn't a solution' I hear you shout. This is true, but think back to when we squared the equation. We have also solved *[tex \sin 2x - \cos 2x = -1] because it squares to the same thing. So you have to check which solutions satisfy your original equation and then you're done.

Hope that helps. (If you do find an easier way to solve it I would be interested to know)

Kev