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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: pillowcase on October 16, 2014, 07:15:06 pm

Title: Methods question help!!!
Post by: pillowcase on October 16, 2014, 07:15:06 pm: can someone please explain this question to me :), thanks in advanced

solve: tan(3x + 3pi)= 1/root3 where x E [-pi, pi]

i got to the first step : 3x+3pi=tan^-1(1/root3) but i havent really got a clue of what to do next, any help would be appreciated :), thanks again

Note: this is exam 1 (non-tech)
Title: Re: Methods question help!!!
Post by: plato on October 16, 2014, 11:53:44 pm: $tan(3x+3\pi) = \frac{1}{\sqrt3}$

Then
$3x+3\pi = \frac{\pi}{6}\pm n\pi$

Given that
$x \epsilon [-\pi,\pi]$ ,
it follows that
$3x \epsilon [-3\pi,3\pi]$
and then
$(3x +3\pi) \epsilon [-3\pi+3\pi,3\pi+\3pi]$
and so
$(3x +3\pi) \epsilon [0, 6\pi]$

Therefore $(3x+3\pi) = \frac{\pi}{6}, \frac{\pi}{6}+\pi, \frac{\pi}{6}+2\pi, \frac{\pi}{6}+3\pi ....... \frac{\pi}{6}+5\pi$
Subtract $3\pi$ throughout.

Then $3x = \frac{\pi}{6}-3\pi, \frac{\pi}{6}-2\pi, \frac{\pi}{6}-\pi, \frac{\pi}{6}....... \frac{\pi}{6}+2\pi$
and now divide by 3 throughout

and so $x = \frac{-17\pi}{18}, \frac{-11\pi}{18}, \frac{\pi}{18}....... \frac{13\pi}{18}$

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