# Solving trig equation for giving range

1. Sep 6, 2008

### thomas49th

1. The problem statement, all variables and given/known data

(b) Solve, for 0 <= x < 360°, the equation
2tan²x + secx = 1,

2. Relevant equations

I think I have to solve it quadratically but I need to get the trigs to the same trig form

3. The attempt at a solution

$$2\frac{sin^{2}x}{cos^{2}x} + \frac{1}{cosx} = 1$$

but I cant see any other logical step after that?

Thanks :)

2. Sep 6, 2008

### Defennder

How do you relate tan^2 x to sec^2 x ?

3. Sep 6, 2008

### HallsofIvy

Staff Emeritus
It would seem logical to me to multiply both sides by cos2 x: 2sin2 x+ cos x= cos2 x. Now replace sin2 x by 1- cos2 x and you have a quadratic equation for cos x.

4. Sep 6, 2008

### tiny-tim

Or, as Defennder was hinting , use one of the standard trigonometric identities :
tan²x = sec²x - 1

Last edited: Sep 6, 2008
5. Sep 6, 2008

### thomas49th

yay ya i got it :) cheers :)