Question

solving for y again...

tan(3y+2)=x^2+5

Answer 1

I have so far y=1/3((x^2+5)/tan-2)

Answer 2

For this question you would have to do a bit of trig. First to get rid of the tan on the left side you have to do the tangent inverse to both sides.

Answer 3

So it would actually be y=(1/3)tan^/1(x^2+5)-2 ?

Answer 4

-1 *

Answer 5

You have the right answer but wrong notation. Parenthesis make a big difference. It should be Y=(tan^-1(x^2+5))/3-2

Answer 6

dont forget the -1 in the exponent. Its the inverse.

Answer 7

but the whole equation would be subtracted by 2 not just (x^2+5)?

Answer 8

Yeah so it should look exactly like this. \[Y=(\frac{ \tan^{-1}(x^2+5)}{ 3})-2\]

Answer 9

Perfect :) thank you again!

Answer 10

No problem! :D Anytime!