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!!! =)
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 $\displaystyle \frac12?$
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:
$\displaystyle x=\frac{120-\pi}{6}-k\frac{\pi}{2}$