Question

Let’s make sure we can handle the symbolic and mechanical aspects of
working with the inverse trigonometric functions:
(a) Find four solutions of tan(2x2 +x−1) = 5.
(b) Solve for x: tan−1
(2x2 + x − 1) = 0.5

Answer 1

@imqwerty, @isry @justjm
I am a bit confused on how to do part A. (this is precalc btw)
For B, is it just
\[\tan^{-1} (2x^2+x-1) = .5\]
\[\tan (\tan^{-1} (2x^2+x-1)=\tan (.5)\]
So
\[2x^2+x-1=\tan (.5)\]
And then
\[(2x-1)(x+1)=\tan(.5)\]
And then individually solve for both equations?
\[(\tan(.5) +1)/2\] and \[\tan(.5) -1\]

Answer 2

@Vocaloid

Answer 3

@jhonyy9

Answer 4

@TheSmartOne

Answer 5

@dude

Answer 6

Rip my work went away

But B is right, solve for each

For A you can try doing the same process but add \(+\pi n\) to the \(tan^{-1}(5)\) because tangent returns to its value every pi units [It's period]

Answer 7

n can be substituted for any value +/- n value

Answer 8

was it my fault? sorry

Answer 9

try using the quadratic formula.

Answer 10

So how I got (tan(.5)+1)/2
and tan(.5)−1, I just add integer of pi to it?
Like (pi+(tan(.5))+1)/2 and pi+tan(.5)−1
And multiply pi by any integer. But for some reason, none of the answer choices are working, the answer key said that sample answers for part a are −1.9293, −1.3677, 0.8677, 1.4293 and I don't know how it is getting those values.

Answer 11

So you want to have

\((2x^{2}+x-1)=tan^{-1}(5)+0\pi\) => \(\color{green}{(2x^{2}+x-1)=tan^{-1}(5)}\)
This will give you 2 solutions

\((2x^{2}+x-1)=tan^{-1}(5)+1\pi\) => \(\color{green}{(2x^{2}+x-1)=tan^{-1}(5)+\pi}\)

You can't really solve this by factoring so you could either complete the square or use the quadratic formula

Answer 12

The second will give you the other two

Answer 13

Wait. We can't factor (2x^2+x−1)?
Doesn't it factor into 2x-1 and x+1?

Answer 14

And we want this in radian mode
Correct?

Answer 15

Yes but the issue is that if you factor it you'll get two totally different answers
You'll get two lines rather than a parabola

And yes radians

Answer 16

It would be easier though quadratic eq

Answer 17

hold on, lemme do that real quick

Answer 18

You'll need a calculator for the decimal

Answer 19

(2x^2+x−1) for this? doesn't it just factor out to 0 and 1/2?

Answer 20

Don't factor
By factoring you alter the equation

\( (2x^2+x−1)=tan^{-1}(5)\)
Set equal to zero
\( 2x^2+x−1-tan^{-1}(5)=0\)

Take the values \((1-tan^{-1}(5))\) as the c value (constant)

You'll end up with something like this

\(\dfrac{-1\pm\sqrt{(1^2-4((2)(1-tan^{-1}(5))))}}{2(2)}\)

Spliting to + and -:

\(\dfrac{-1+\sqrt{(1^2-4((2)(1-tan^{-1}(5))))}}{2(2)}\) and \(\dfrac{-1-\sqrt{(1^2-4((2)(1-tan^{-1}(5))))}}{2(2)}\)
This is so long thats why it would be better to solve with calc


======================================


For the other equation:
\( (2x^2+x−1)=tan^{-1}(5)+\pi\)
Set equal to zero
\( 2x^2+x−1-tan^{-1}(5)+\pi=0\)

Take the values \(1-tan^{-1}(5)+\pi\) as the constant




Now for this one you would do a similar one as previous only adding pi to it


\(\dfrac{-1+\sqrt{(1^2-4((2)(1-tan^{-1}(5)+\pi)))}}{2(2)}\) and

\(\dfrac{-1-\sqrt{(1^2-4((2)(1-tan^{-1}(5)+\pi )))}}{2(2)}\)

Answer 21

Quadratic eq is easier to use if they give you quadratic equations


Treat the trig function as a constant if they do not have x

Answer 22

Thanks @dude, that explanation really helped. You are Great!!!

Answer 23

But dude, The equations where you added pi to it, are undefined. What does that mean?

Answer 24

Hmm
What calculator are you using?

I think some manual ones use arctan instead of tan^-1

Answer 25

Or the equation might be too long

Answer 26

works for me... im using ti 84 plus

Answer 27

try redoing the negatives... maybe you did a negative instead of a minus symbol.

Answer 28

anyway quadratic formula did work though...

Answer 29

I used degrees but radians still should work... it does for me

Answer 30

???

Answer 31

Use radians D:

Answer 32

Still doesn't work. Could you screenshot what you did?

Answer 33

Oh I goofed

I missed the negative sign in front of the 1 and pi

First two set should be:
\(\dfrac{-1+\sqrt{1^2-4(2)(\color{red}{-}1-tan^{-1}(5))}}{2(2)}\) and \(\dfrac{-1+\sqrt{1^2-4(2)(\color{red}{-}1-tan^{-1}(5))}}{2(2)}\)

Second set should be:
\(\dfrac{-1+\sqrt{1^2-4(2)(\color{red}{-}1-tan^{-1}(5)\color{red}{-}\pi)}}{2(2)}\) and \(\dfrac{-1+\sqrt{1^2-4(2)(\color{red}{-}1-tan^{-1}(5)\color{red}{-}\pi)}}{2(2)}\)

Answer 34

Oh lol, its okay. thanks dude.