Solving trig equation for giving range

[thomas49th]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

but I can't see any other logical step after that?

Thanks :)

 

[Defennder]

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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

but I can't see any other logical step after that?

Thanks :)

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.

 

[tiny-tim]

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.

Or, as Defennder was hinting , use one of the standard trigonometric identities :

tan²x = sec²x - 1 ​

 

[thomas49th]

yay you i got it :) cheers :)