Question

use the inverse functions where necessay to find all the solutions of the equation in the interval [0, 2pi)

tan^2 x + tan x - 12 = 0

Answer 1

First, you need to notice that this is merely a quadratic equation where \(u=\tan(x)\). Can you factor and solve \[u^2+u-12=0?\]

Answer 2

i tried but then iam stuck

Answer 3

tan x (tanx + 1) =12

Answer 4

Possible factorizations for -12 are \(\pm(1, 12), \pm(2, 6), \pm(3, 4)\) Notice that if you choose \(-3, 4\), you get a factorization of \[(u-3)(u+4)\]Thus, your original equation is the same as \[(\tan x+4)(\tan x-3)=0\]

Answer 5

This means that either \(\tan x=-4\) or \(\tan x=3\). With these, use the inverse function to find these.

Answer 6

yes i get it thanks

Answer 7

Also, be careful about the domain of \(\tan x\). You need to give your solutions in the interval \([0, 2\pi]\), but the inverse tangent function will give you values in the interval \([-\pi/2, \pi/2]\), so you need to add multiples of \(\pi\) until you have all solutions in \([0, 2\pi]\).

Answer 8

sorry to bother u but just a secound

i checked my answer in the back of the book and it is

1.2490, 1.8158, 4.3906, 4.9574

i got the first answer but where did the rest of the answers come form?

Answer 9

From what I said directly above. If you add \(\pi\) to 1.2490, you should get 4.3906. Also, try adding \(\pi\) and \(2\pi\) to the answer you got for \(\tan^{-1}(-4)\). That should get you the other two solutions.

Answer 10

ok so how do i know how many answers it is supposed to

Answer 11

You know that the inverse function for tan gives values from \([-\pi/2, \pi/2]\). Note that this has length \(\pi\). This means that \[\tan^{-1}(x)=\tan^{-1}(x)+\pi=\tan^{-1}(x)+2\pi=...\]Now your interval \([0, 2\pi]\) has length \(2\pi\), so that means \(\tan^{-1}(x)\) has 2 solutions in that interval.

You have two inverse tangents to calculate, so you have 4 solutions total.

Answer 12

ok

Answer 13

Did that make sense?

Answer 14

a little is there is an easier way to know it or at least another way

Answer 15

it is ok i got it thanks

Answer 16

I suppose there's always guess and check. You still need to know that \[\tan^{-1}(x)=\tan^{-1}(x)+\pi=\tan^{-1}(x)+2\pi=...\]So you get that \(\tan^{-1}(-4)\approx-1.32582\). Now you check if this is in the interval \([0, 2\pi]\). It isn't. So add \(\pi\). You get about 1.8158. Is this in the interval? Sure is!

Now we keep adding another \(\pi\) until we exit the interval. \(1.8158+\pi\approx4.9574\) which is still in the interval. However, if we add another \(\pi\), it's out of the interval, so we stop there.

Answer 17

yes i get it thankyou so much

Answer 18

You're welcome.