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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: samsiexD on April 25, 2013, 03:47:22 pm

Title: Finding exact solutions
Post by: samsiexD on April 25, 2013, 03:47:22 pm: Hey so i'm stuck on a question i did one step but didnt know where to go next :P

Find the exact solution to:
Sin(4x)=cos(2x) for xE[-pi,pi]
I did
2sin(2x)cos(2x)=cos(2x) but didn't know what to do next

Help is appreciated :)
Title: Re: Finding exact solutions
Post by: brightsky on April 25, 2013, 03:52:49 pm: so subtract cos(2x) from both sides of the equation, factorise the cos(2x) out and then use null factor law to find the solutions.
Title: Re: Finding exact solutions
Post by: Jeggz on April 25, 2013, 03:53:42 pm: So continuing on from yourself -
$2sin(2x)cos(2x)=cos(2x)$

$2sin(2x)cos(2x)-cos(2x)=0$

$cos(2x)[2sin(2x)-1]=0$

Now you have two equations that you can solve. The first one being $cos(2x)=0$ and the second one being $2sin(2x)-1=0$

Hope that helps :)
Title: Re: Finding exact solutions
Post by: samsiexD on April 25, 2013, 04:01:32 pm: Okies, thanks to both :)