Question

The value of pi/4 is a solution to the equation
3 sqrt 2 sec theta+7 = 1

Answer 1

this is true or false. thus far I tried to solve the equation and I got
sec theta = sqrt 2/2

Answer 2

sec theta = sqrt2/2 ?
Hmm that doesn't look quite right :o

Answer 3

\[\large\rm 3\sqrt2 ~\sec \theta+7=1\]Subtracting 7 gives,\[\large\rm 3\sqrt2 ~\sec \theta=-6\]Dividing by 3sqrt2 gives,\[\large\rm \sec \theta=\frac{-6}{3\sqrt2}\qquad\to\qquad \sec \theta=-\frac{2}{\sqrt2}\]Something more like that, ya? :)

Answer 4

I took away 1 on each side.
Then I divided both sides by 3
then by root 2
so actually in this case it would be sec theta = 2/sqrt2

Answer 5

Oh yeah I see where that is the better answer and the negative shouldve carried over in my answer. okay

Answer 6

Divide 2 by sqrt2 to get,\[\large\rm \sec \theta=-\sqrt 2\]

Wait wait wait, why all this fancy business?
Why not just plug pi/4 into the equation and see if it holds true? :)

Answer 7

very true. let me try that. thanks for pointing that out!

Answer 8

wait pi/4 replaces theta right?

Answer 9

yes

Answer 10

also i dont know to to enter sec into my calculator, there is no sec button

Answer 11

\[\large\rm 3\sqrt2 ~\color{orangered}{\sec \frac{\pi}{4}}+7=1\]
\[\large\rm 3\sqrt2 ~\color{orangered}{\frac{1}{\cos\frac{\pi}{4}}}+7=1\]You don't need to use a calculator anyway you silly billy! >.<
Don't you remember your unit circle and those special values? :)

Answer 12

Yeah I forgot that sec is 1/cos

Answer 13

but I dont see how you can solve without a calculator

Answer 14

You haven't learned special values yet? :o

cosine of pi/4 = sqrt(2)/2, yes?

Answer 15

yes i know this. itis on the unit circle. I am not able to do those things without using the unit circle yet

Answer 16

so where ever cos is equal to pi/4 which is sqrt2/2

Answer 17

therefore it is a solution

Answer 18

Woah woah, hold on. You're jumping too far ahead.
\[\large\rm 3\sqrt2 ~\color{orangered}{\frac{1}{\cos\frac{\pi}{4}}}+7=1\]Yes, cos(pi/4) is sqrt2/2,
\[\large\rm 3\sqrt2 ~\color{orangered}{\frac{1}{\left(\frac{\sqrt2}{2}\right)}}+7=1\]

Answer 19

Now simplify further.
Remember how to deal with `division by a fraction`?

Answer 20

No fractions are not my forte at all

Answer 21

Remember your uhh, what is it that they teach you rascals nowadays.. um
Keep Change Flip? :)

Answer 22

I think I know that saying, yeah

Answer 23

\[\large\rm \frac{1}{\sqrt2/2}=1\cdot\frac{2}{\sqrt2}\]
`Keep` the numerator the same
`Change` the operation from division to multiplication
`Flip` the bottom fraction.

Answer 24

So our orange mess turns into this:\[\large\rm 3\sqrt2 \cdot\color{orangered}{\frac{2}{\sqrt2}}+7=1\]

Answer 25

I see, that wasnt so bad but it will probably take time for me to catch on. now what though?

Answer 26

Keep simplifying, see whether or not the equality holds true.

Answer 27

I would probably subtract the 7 again,\[\large\rm 3\sqrt2\cdot\frac{2}{\sqrt2}=-6\]And then simplify the left side to see if this holds true.

Answer 28

okay i get 2/sqrt2=2/sqrt2

Answer 29

i mean 2/sqrt2=-2/sqrt2

Answer 30

Which is NOT true, ya? :)

Answer 31

yeah so this means the answer is false that pi/4 is not a solution

Answer 32

Yay good job \c:/

It turns out that 3pi/4 and 5pi/4 are solutions,
but not pi/4.

Answer 33

Cool thanks for your help. I will probably have a few more questions tonight so if you'd like to join dont hesitate. I am taking a test and i get stuck sometimes

Answer 34

cool ୧ʕ•̀ᴥ•́ʔ୨