trig equation

Nancy · #1 $Old$ March 21st, 2009, 02:57 PM

Hi!
I know this is a fairly simple trig equation but i just keep getting it wrong!

equation:
tan(40-2x)=3^1/2 (root 3)

Can someone please help me?

Thank u!!! =)

Reckoner · #2 $Old$ March 21st, 2009, 04:15 PM

Quote:

Originally Posted by Nancy $View Post$

Hi!
I know this is a fairly simple trig equation but i just keep getting it wrong!

equation:
tan(40-2x)=3^1/2 (root 3)

You should provide your work so that we can point out any mistakes.

Begin by noting that $\sqrt3=\frac{2\sqrt3}2=\frac{\sqrt3/2}{1/2}.$ The tangent is sine divided by cosine, so for what angles does sine equal $\frac{\sqrt3}2$ and cosine equal $\frac12?$

controlfreddy · #3 $Old$ March 22nd, 2009, 07:50 AM

you need to calculate when the argument of the tan is
equal to $\sqrt{3}$
Remembering that $\tan{x} = \sqrt{3}$ when
$x=\frac{\pi}{3}+k\pi$ as also Reckoner said.

Then u can write:

$40-2x= \frac{\pi}{3}+k\pi$

$2x=-\frac{\pi}{3}-k\pi+40$

$x=-\frac{\pi}{6}-k\frac{\pi}{2}+20$

simplifying:

$x=\frac{120-\pi}{6}-k\frac{\pi}{2}$

Nancy · #4 $Old$ March 25th, 2009, 06:03 AM

Thank u! for ur help....i figured it out....