trigonometric functions

[fiery123]

2(tanx-1)^2 = 3(sec^2(x) -2)

[VBAhack]

What's the question?

[fiery123]

whoops its solving the equation for 0<=x<=360

[fourtonmantis]

Hey there. I assumed you were trying to solve for x and I went ahead and had a crack at this:

2(tan(x) - 1)^2 = 3(sec^2(x) - 2)
2(tan(x) - 1)^2 = 3(tan^2(x) + 1 - 2)
2(tan(x) - 1)^2 = 3(tan^2(x) - 1)
2*tan^2(x) - 4*tan(x) + 2 = 3*tan^2(x) - 3
-tan^2(x) - 4*tan(x) + 5 = 0
tan^2(x) + 4*tan(x) - 5 = 0
(tan(x) + 5)(tan(x) - 1) = 0
tan(x) = -5 or tan(x) = 1

so x = arctan(-5) or x = arctan(1) = pi/4.

Hope that helps .

[fiery123]

The answer in the book actually says x = 63.4 or 243.4.
But since i got the same answer as you previously i think the book is wrong, ok thanks.