Solve giving all solutions in the interval 0 ≤ x ≤ 180

**johnreal** · March 18th 2012, 08:23 AM

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

**skeeter** · March 18th 2012, 08:33 AM

Originally Posted by johnreal

I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 ...

$\sec{x} = 4 - 2(\sec^2{x} - 1)$

$\sec{x} = 4 - 2\sec^2{x} + 2$

$2\sec^2{x} + \sec{x} - 6 = 0$

this quadratic will factor ...

**princeps** · March 18th 2012, 08:42 AM

Originally Posted by johnreal

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

$\frac{1}{\cos x}=4-2\cdot \frac{\sin^2 x}{\cos^2 x} \Rightarrow \frac{1}{\cos x}=4-2\cdot \frac{1- \cos^2 x}{\cos^2 x} \Rightarrow$

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

Now substitute :

$\cos x=t$

and solve quadratic equation .

**Plato** · March 18th 2012, 08:42 AM

Originally Posted by johnreal

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

Rewrite the equation: $2\sec^2(x)+\sec(x)-6=0$ .
The solve $2u^2+u-6=0\text{ where }u=\sec(x)~.$

**johnreal** · March 18th 2012, 09:53 AM

Thanks all of you, the two different methods really helped me understand and the substitution steps cleared it all up, thank alot guys.

**johnreal** · March 18th 2012, 10:38 AM

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.

**johnreal** · March 18th 2012, 10:47 AM

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.

**highvoltage** · March 18th 2012, 01:11 PM

Oops! I did this and realized you already have the solution. At any rate here it is.

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Solve giving all solutions in the interval 0 ≤ x ≤ 180

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Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

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