Question

solve the given equation in term of 2kpi
tan^2(x)-2sec(x)=2

Answer 1

We need to put this in terms of one trig function
Recall:
\[1+\tan^2(x)=\sec^2(x)\]
=>
\[\tan^2(x)=\sec^2(x)-1\]
So we have the equation
\[\tan^2(x)-2\sec(x)=2\]
To put in terms of sec(x) we will use the identity above
\[(\sec^2(x)-1)-2\sec(x)=2\]
So we have
\[\sec^2(x)-1-2\sec(x)=2\]
Now subtracting 2 on both sides and reordering the terms we have
\[\sec^2(x)-2\sec(x)-1-2=0\]
Combining like terms gives
\[\sec^2(x)-2\sec(x)-3=0\]
Now do you know how to solve:
\[u^2-2u-3=0\]
If you know how to solve that equation then you will be able to solve the first since these equations have the relationship
\[u=\sec(x)\]
So tell we what you get when you solve
\[u^2-2u-3=0\]
And then we will go from there.

Answer 2

are you using calculus to solve this?

Answer 3

No

Answer 4

I used a trig identity

Answer 5

so....(secx-2)=0 (secx+1)=0???

Answer 6

No

Answer 7

\[u^2-2u-3=0\]
Try factoring this first
You will get
(sec(x)-3)(sec(x)+1)=0

Answer 8

since -3(1)=-3 and -3+1=-2

Answer 9

Now set both factors =0

Answer 10

secx=3 secx=-1

Answer 11

Right

Answer 12

the answer in the book doesnt make sense

Answer 13

Why do you say that?

Answer 14

because its in decimal form, I'm confused on when to use exact form and decimal. or does that not even matter?

Answer 15

It does matter
What does the question say?
Does it say you can round?

Answer 16

Usually it says what to do

Answer 17

It shouldn't be a guessing name
If it doesn't say anything about rounding then you provide the exact answer

Answer 18

solve the given equation

Answer 19

gosh game*

Answer 20

sorry

Answer 21

in the book the answers for that area jump from decimal to exact

Answer 22

ok if it doesn't say anything about rounding i would put the exact answer regardless of what the back of the book says

Answer 23

ok

Answer 24

the answers are: (2k+1)pi, 1.23+2kpi, 5.05+2kpi

Answer 25

Those are approximations though

Answer 26

well the last two

Answer 27

Also confused on how they got to that point

Answer 28

ok so we have
sec(x)=3 or sec(x)=-1

Answer 29

ok

Answer 30

Or we could say
since \[\sec(x+2 k \pi) = \sec(x)\]
then
\[\sec(x+ 2 k \pi)=3 \text{ or } \sec(x+2 k \pi ) =-1\]

Answer 31

Now take arcsec( ) of both sides

Answer 32

\[=> x+2 k \pi = \sec^{-1}(3) \text{ or } x+2 k \pi =\sec^{-1}(-1)\]

Answer 33

oh is that where the approximations come from?

Answer 34

yep i assume they approximate arcsec(3) but arcsec(-1) you can evaluate that exactly

Answer 35

But....

Answer 36

ok now I think I understand.

Answer 37

Since cos(x) is an even function then sec(x) is an even function
We do have more solutions

Answer 38

\[\sec(x)=\sec(-x)\]

Answer 39

So we have the above+
\[x+2k \pi=-\sec^{-1}(3) \text{ or } x+2 k \pi=-\sec^{-1}(-1)\]

Answer 40

so problems like this i would break down the first part of the equation to match the second part and take the arc of the identity?

Answer 41

Sometimes and also you might need to use the fact that a trig function is odd or even like i did

Answer 42

Ok. I'm going to try a few problems thanks

Answer 43

Np
So I gave you 4 equations that will give you your solution just so you know :)