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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: zibb3r on April 27, 2011, 10:04:46 pm

Title: Solve this equation????
Post by: zibb3r on April 27, 2011, 10:04:46 pm: Solve sin(2x) + 2sin(2x)cos(2x) = 0 where x is [-pi,3pi]

Not sure, any idea how????

THanks
Title: Re: Solve this equation????
Post by: xZero on April 27, 2011, 10:10:27 pm: $sin(2x)+2sin(2x)cos(2x)=0$

$sin(2x)(1+2cos(2x))=0$ , where $-2\pi\leq 2x \leq 6\pi$

solve for 2x within the restriction and divide both side by 2 to find x
Title: Re: Solve this equation????
Post by: zibb3r on April 27, 2011, 10:37:15 pm: Would you e able to explain further????? Im still a bit confused :(
Title: Re: Solve this equation????
Post by: taiga on April 27, 2011, 10:41:28 pm: You can then break
$sin(2x)(1+2cos(2x))=0$

into

$sin(2x)=0$ and $(1+2cos(2x))=0$

From there you should be able to solve it :)

If you can't (I'm not saying this in a mean way) then you should check out some of your earlier maths books :)
Title: Re: Solve this equation????
Post by: xZero on April 27, 2011, 10:48:51 pm: ^this, looking through books and working it out yourself will be more beneficial than us posting the solution. Of course if you can do it then ignore this :)
Title: Re: Solve this equation????
Post by: zibb3r on April 27, 2011, 10:49:36 pm: I can do that and the null factor law just confused about what to do next.......

My text book is at school aswell.....
Title: Re: Solve this equation????
Post by: xZero on April 27, 2011, 10:54:20 pm: try replacing 2x with u and solve for u, in this case it will be sin(u)=0 or 1+2cos(u)=0. (remember the domain of 2x)
Title: Re: Solve this equation????
Post by: zibb3r on April 27, 2011, 10:56:46 pm: Nevermind, thanks to all. I guess trig just isnt my thing! Will leave it:D
Title: Re: Solve this equation????
Post by: brightsky on April 28, 2011, 07:47:14 pm: sin(2x) + 2 sin(2x) cos(2x) = 0
sin(2x) (1 + 2 cos(2x)) = 0
sin(2x) = 0 or 1 + 2cos(2x) = 0
from sin(2x) = 0, we get:
2x = -2pi, -pi, 0, pi, 2pi, 3pi, 4pi, 5pi, 6pi [remember restriction]
x = -pi, -pi/2, 0, pi/2, pi, 3pi/2, 2pi, 5pi/2, 3pi
from 1 + 2cos(2x) = 0
we get cos(2x) = -1/2
so:
2x = -4pi/3, -2pi/3, 2pi/3, 4pi/3, 8pi/3, 10pi/3, 14pi/3, 16pi/3
x = -2pi/3, -pi/3, pi/3, 2pi/3, 4pi/3, 5pi/3, 7pi/3, 8pi/3
so we have a total of...17 solutions...