Expert: Ahmed Salami - 4/14/2010

Question
I have tried these problems and can't seem to factor them out.

Solve the following equations given that 0<=x<360
a) 2tan^2 2x=3sec2x
b) cot^2 x=3

Answer
Hi Drew,
a) Firstly, by identity
tan²(2x) = sec²(2x) - 1
So,
2tan²(2x) = 3sec(2x)
becomes
2[sec²(2x) - 1] = 3sec(2x)
2sec²(2x) - 2 = 3sec(2x)
2sec²(2x) - 3sec(2x) - 2 = 0
Rearranging and factorizing,
2sec²(2x) - 4sec(2x) + sec(2x) - 2 = 0
2sec(2x)[sec(2x) - 2] + [sec(2x) - 2] = 0
[sec(2x) - 2][2sec(2x) + 1] = 0
and we have
sec(2x) - 2 = 0    
sec(2x) = 2       
cos(2x) = 1/2      
2x = 60°, 300°, 420°, 660°
x = 30°, 150°, 210°, 330°

OR
2sec(2x) + 1 = 0
2sec(2x) = -1
sec(2x) = -1/2
cos(2x) = -2
and of course this has no solution since cos(2x) cannot be greater than 1 or less than -1.

b) cot²x = 3
tan²x = 1/3
tanx = 1/√3   OR   tanx = -1/√3
x = 30°, 210°   OR   x = 150°, 330°
x = 30°, 150°, 210°, 330°


Regards