Help with a trig question?

Question

solve the equation sec x = -2 in the interval 0 ≤ x ≤ pi

Any help would be appreciated. I used to know trig very well but haven't done it in a while. Please explain as detailed as possibly so that I am able to do similar problems on my own. Thanks so much and 10 points to the best answer.

Answer 1

First use the trig identity

sec x = 1/cos x

1/cos x = -2

cos  x = -1/2

Now use the special triangle to solve for that

http://fouss.pbwiki.com/f/special%20tria...

SOH CAH TOA

Cos x =  adjacent/hyp

cos^-1 1/2 = pie/3

and -, + you could just pie and bring cos in the -negative quadrant  :)

The short form to that is

"All students take calculus"

Hope tht helps :)

Answer 2

sec x = -2 in the interval 0 ≤ x ≤ pi

We know, that cos x = 1/sec x; hence,

cos x = - 1/2

Since the sign of cos is negative and x is 0 ≤ x ≤ pi, then x is in the 2nd quadrant.

x = arcos(-1/2)

x = 2.094 radians or 120deg   ANSWER

Hope this helps.

teddy boy

Answer 3

Well, remember that secant is 1/cos, so if sec x is -2, then

cosx = -1/2.  Since you know that x has to be between 0 and pi, then the angle has to be in the second quadrant because that is where the x-values (ie. the cosines) are negative.  So then it is just a matter of remembering which one of your special angles gives x-values of 1/2, and it is 2*pi/3 or 120 degrees.  Hope that helps!  =)