cosine problem

**jonathanraxa** · May 8th 2012, 03:38 PM

" cos(x^2+5x+9)

round answer to 4 decimal places. "
Thanks!

**skeeter** · May 8th 2012, 04:14 PM

Originally Posted by jonathanraxa

" cos(x^2+5x+9)

round answer to 4 decimal places. "
Thanks!

what do you want done w/ this expression? directions?

**jonathanraxa** · May 8th 2012, 04:27 PM

Originally Posted by skeeter

what do you want done w/ this expression? directions?

Sorry I forgot to mention. I'm trying to solve for 'x'

**skeeter** · May 8th 2012, 05:21 PM

Originally Posted by jonathanraxa

Sorry I forgot to mention. I'm trying to solve for 'x'

then it would help to know ...

$\cos(x^2+5x+9) = \, ?$

**jonathanraxa** · May 8th 2012, 05:23 PM

Originally Posted by skeeter

then it would help to know ...

$\cos(x^2+5x+9) = \, ?$

sorry again. It would equal 0

**Plato** · May 8th 2012, 05:31 PM

Originally Posted by jonathanraxa

sorry again. It would equal 0

That means that $x^2+5x+9$ must be an odd multiple of $\frac{\pi}{2}$ .

**skeeter** · May 8th 2012, 05:31 PM

if ...

$\cos(x^2+5x+9) = 0$

then ...

$x^2 + 5x + 9 = \frac{(2k+1)\pi}{2}$

$k \in \mathbb{Z}$

**jonathanraxa** · May 8th 2012, 09:28 PM

Another thing I forgot to mention was that I'm trying to find the smallest positive number 'x' from this equation. I don't know if this helps in any way

**skeeter** · May 9th 2012, 03:20 AM

Originally Posted by jonathanraxa

Another thing I forgot to mention was that I'm trying to find the smallest positive number 'x' from this equation. I don't know if this helps in any way

Plato and I have laid the groundwork for a solution ... you figure out the specific solution using this last "forgotten" detail.

Next time, try to post the entire problem as written.

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