Question

How in the world do I solve this?
sec^2 (x)+sec (x)-2=0
I know you have to simlify it so I added two on both sides so far and I am stuck after that

Answer 1

its a disguised quadratic equation

Answer 2

do you know your quadratic formula?

Answer 3

yes

Answer 4

but right now I am learning the basics of solving this and I know that you must first simplify

Answer 5

if you are learning the basics ,,, then complete the square, which is just the proofing of the quad formula

Answer 6

in this case, id use the quad formula if you are expected to already know it.

Answer 7

How would this plug into the quadratic formula?

Answer 8

let u=sec(x)

u^2 +u - 2 = 0

what is your formula?

Answer 9

a=1
b=1
c=-2

Answer 10

good

Answer 11

-1+-sqrt 1^2-4(1)(-2) all of that over 2

Answer 12

but then you get a negative in the sqrt

Answer 13

therefore:
\[u=sec(x)=\frac{-1\pm\sqrt{1-4(1)(-2)}}{2(1)}\]

recall that sec = 1/cos soo

cos(x)=2/(1 +- sqrt(9))

x = cos^(-1) (....)

Answer 14

no, no you dont :)

Answer 15

1 +- 3 = -2,4

soo

cos(x) = -1, 1/2 is our solutions

Answer 16

oh yeah i forgot... so it would be the sqrt of 9 which is 3
so -1+3/2 and -1-3/2

Answer 17

if you go with sec yes,

Answer 18

so 1 and -2

Answer 19

sounds good to me

Answer 20

arc trig them to get x

Answer 21

but the thing is there are three solutions that need to be found and that is only two

Answer 22

thats not the end of it ... when you inverse the trig you will sovle for x

Answer 23

\[sec(x)=\frac{1\pm3}{2}\]

Answer 24

opps, forgot the 1 is negtaive

Answer 25

you still get 1, and -2

Answer 26

\[sec(x)=\frac{-1\pm3}{2}\]
\[sec(x)=-2,1\]
\[cos(x)=1,-\frac{1}{2}\]
cos(x)=-1/2 has 2 values between 0 and 2pi
cos(x)=1 only has 1 value