I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 , i have no idea where to start..... so far i changed the tan into sec^2-1, but i still cannot solve, i think they want me to use binomial expansion (Headbang)

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- Mar 18th 2012, 08:23 AMjohnrealSolve giving all solutions in the interval 0 ≤ x ≤ 180
I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 , i have no idea where to start..... so far i changed the tan into sec^2-1, but i still cannot solve, i think they want me to use binomial expansion (Headbang)

- Mar 18th 2012, 08:33 AMskeeterRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
- Mar 18th 2012, 08:42 AMprincepsRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
$\displaystyle \frac{1}{\cos x}=4-2\cdot \frac{\sin^2 x}{\cos^2 x} \Rightarrow \frac{1}{\cos x}=4-2\cdot \frac{1- \cos^2 x}{\cos^2 x} \Rightarrow$

$\displaystyle \Rightarrow \cos x =4\cos^2 x-2+2\cos^2 x \Rightarrow 6\cos^2 x-\cos x-2=0$

Now substitute :

$\displaystyle \cos x=t$

and solve quadratic equation . - Mar 18th 2012, 08:42 AMPlatoRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
- Mar 18th 2012, 09:53 AMjohnrealRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
Thanks all of you, the two different methods really helped me understand and the substitution steps cleared it all up, thank alot guys.

- Mar 18th 2012, 10:38 AMjohnrealRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
ok guys one last clarification, i now have sec x=3/2 and sec x=-2, how can i find for x.... this is the first time i'm doing these kind of questions.

- Mar 18th 2012, 10:47 AMjohnrealRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
Never mind i got it, i put sec as 1/cos then reversed it so cos=2/3 and cos=-1/2 which resulted in 48.2 and 120 degrees, thank again i think im getting the hang of this.

- Mar 18th 2012, 01:11 PMhighvoltageRe: Solve giving all solutions in the interval 0 ≤ x ≤ 180
Oops! I did this and realized you already have the solution. At any rate here it is.

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