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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: naved_s9994 on November 02, 2009, 11:05:01 am

Title: Trig
Post by: naved_s9994 on November 02, 2009, 11:05:01 am: Heffernan Exam 1 2006 Q4 Solutions. Solving for x

Can anyone verify the exam's solutions'.
Also, how would you type this into CAS, to check (Although this is an Exam 1?)

Solutions to the Question Attached.

Edit: Because it is a VERY simple question, and getting it wrong is very disturbing!
Title: Re: Trig
Post by: m@tty on November 02, 2009, 11:09:52 am: The solutions are right, and I don't do CAS

Where did you go wrong?
Title: Re: Trig
Post by: naved_s9994 on November 02, 2009, 11:13:14 am: I did it and got x= 2pi/3 and 5pi/6

sin(2x) = -sqrt(3)/2
Hence 2x=-pi/3

and since its negative.....your looking in 3rd and 4th quadrant...

therfore pi+ pi/3
and 2pi-pi/3

and hmm?
Title: Re: Trig
Post by: m@tty on November 02, 2009, 11:25:13 am: The domain is $x \in [-\frac{\pi}{2},\frac{\pi}{2}]$
$\therefore 2x \in [ - \pi, \pi]$

Since it is negative you are looking in third and fourth quadrant, but you can't get there anti-clockwise(positive) so you need to go clockwise(negative). Thus you arrive at the solutions;
$2x = 0 - \frac{\pi}{3} , -\pi + \frac{\pi}{3}$
$2x= - \frac{\pi}{3} , - \frac{2 \pi}{3}$

$So: x= - \frac{\pi}{6} , - \frac{\pi}{3}$
Title: Re: Trig
Post by: naved_s9994 on November 02, 2009, 11:28:22 am: Thanks a lot! (I got it now), it seems i had forgotten the basics [!?@#$@!~!!]