Question

tan2 x + tan x − 12 = 0

Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)

Answer 1

Screenshot of the math problem>

Answer 2

Heyyy, looks like the above equation can be easily factored! Can you try factoring it?

Answer 3

Screenshot of example of what the answer is supposed to look like

Answer 4

I am confused on how to factor it because I know 1 x -30 = -30(The last term). But the number that are supposed to be added together have to equal 1!

tan2 x + tan x − 12 = 0
(tan -1 )(tan +30 )

Answer 5

Not quite right! This would be the factored form\[\huge(\tan(x)+4)(\tan(x)-3)\]

Answer 6

I'm sorry! I accidentally copied the example from the screenshot on my paper!

Answer 7

I was using tan2 x + tan x − 30 = 0 lol

Answer 8

How would you factor that though?^

Answer 9

Use +6 and -5

Answer 10

\[\large (\tan x+6)(\tan x -5)\]

Answer 11

\[\tan ^2x+\tan x-12=0\]
\[\tan ^2 x+4\tan x-3\tan x-12=0\]
\[\tan x(\tan x+4)-3(\tan x+4)=0\]
\[\left( \tan x+4 \right)\left( \tan x-3 \right)=0\]
\[tanx=-4\]
Hence x lies in 2nd and 4th quadrant.
\[x=\pi- \tan^{-1} (4),2 \pi-\tan^{-1}( 4)\]
\[\tan x=3, x~ lies~ \in~ 1st~ and~ 3rd~quadrant.\]
\[x=\tan^{-1} (3),\pi+\tan^{-1} (3)\]

Answer 12

^ nice

Answer 13

Thank you! Sorry I disappeared for a second!

Answer 14

yw