1. ## find x

hi i am stuck on a question ..pls can you help:

tanx(squared) + 3tanx + 2 = 0

find x:

i got (tanx +1)(tanx+2) = 0

tanx = -1, tanx = -2

2. ## Re: find x

Originally Posted by janvi anish
tanx(squared) + 3tanx + 2 = 0
find x:
i got (tanx +1)(tanx+2) = 0
tanx = -1, tanx = -2
If the question is $\displaystyle \tan^2(x) + 3\tan(x) + 2 = 0$ then yes so far so good.
But the you posted it, it reads $\displaystyle \tan(x^2)$. Learn to post correctly.

Now you can use the $\displaystyle \arctan\text{ function }.$

3. ## Re: find x

Looks good, so far. For the solution tan(x) = -1 you now need to determine the angle x where tan(x) = -1. Hint - it's where sin(x) = -cos(x), and could be in quadrant 2 or 4. For tan(x) = -2 it's little more complicated - again it's in quadrant 2 or 4, but you'll have to find the angle using arctangent(-2).

4. ## Re: find x

thanks
what is arctan? is it tan-1 (like to find angles on the calcualtor or shift tan)
i know that the answer is -45 and 136 because when you use -1 and tan-1 it, those are the answers but and put it into the original equation it equals 0 so it works but i dont understand how -2 works. when you tan-1 it, the answer is -63... and 117... so when you substitute that into the original equation it doesnt equal 0 so im not sure why -1 works and why -2 doesnt?

5. ## Re: find x

Yes, arctangent(x) is often written as $\displaystyle tan^{-1}x$. You have the correct answers -- -45 and 115 degrees (not 136) are equal to arctan(-1), and -63.4 and 116.6 degrees equal arctan(-2). When you put these values into the original equation it should work (it does for me):

$\displaystyle tan^2(116.6) + 3 tan(116.6) + 2 = (-2)^2 +3(-2) + 2 = 0$

$\displaystyle tan^2(-63.4) + 3 tan(-63.4) + 2 = (-2)^2 +3(-2) + 2 = 0$

6. ## Re: find x

okay! thank you for your help!!!

7. ## Re: find x

Correction! I made a typo that I can't fix. This:

Originally Posted by ebaines
... -45 and 115 degrees (not 136) ...
should have been: -45 and 135 degrees.