Question

Find all solutions in the interval [0, 2π).

sec2x - 2 = tan2x

No solution

x = pi/3

x = pi/4

x = pi/6

Answer 1

I'm afraid of the sec and csc functions, so first I will substitute sec 2x = 1/cos2x and tan2x = sin2x/cos2x:\[\frac{ 1 }{ \cos 2x }-2=\frac{ \sin2x }{ \cos 2x }\]Multiply everything with cos2x:\[1-2\cos 2x=\sin 2x\]Now convert sin 2x to cos2x, using the formula sin²x + cos²x = 1:\[1-2\cos 2x=\pm \sqrt{1-\cos^2 2x}\]Squaring gets rid of the root-sign, but will also give false solutions if we don't pay some attention.
First, square both sides:\[1-4\cos 2x+4\cos^2 2x=1-\cos^2 2x\]This is the same as:\[5\cos^2 2x - 4\cos 2x = 0\]Factoring gives:\[\cos 2x(5\cos 2x-4)=0\]Split up:\[\cos 2x = 0\]\[\cos2x=\frac{ 4 }{ 5 }\]
But: cos 2x = 0 doesn't work, because of the original dividing by it (see formula at the top).
So we're left with cos 2x = 4/5.

We have come a long way...
Maybe you want to do the rest yourself first? I hope this helps a bit!

Answer 2

BTW, the solutions I get are none of the above mentioned by you...