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!!! =)

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- Mar 21st 2009, 01:57 PMNancytrig equation
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!!! =) - Mar 21st 2009, 03:15 PMReckoner
You should provide your work so that we can point out any mistakes.

Begin by noting that $\displaystyle \sqrt3=\frac{2\sqrt3}2=\frac{\sqrt3/2}{1/2}.$ The tangent is sine divided by cosine, so for what angles does sine equal $\displaystyle \frac{\sqrt3}2$ and cosine equal $\displaystyle \frac12?$ - Mar 22nd 2009, 06:50 AMcontrolfreddy
you need to calculate when the argument of the tan is

equal to $\displaystyle \sqrt{3}$

Remembering that $\displaystyle \tan{x} = \sqrt{3}$ when

$\displaystyle x=\frac{\pi}{3}+k\pi$ as also Reckoner said.

Then u can write:

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

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

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

simplifying:

$\displaystyle x=\frac{120-\pi}{6}-k\frac{\pi}{2}$ - Mar 25th 2009, 05:03 AMNancy
Thank u! for ur help....i figured it out....

(Rofl)