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anonymous
 one year ago
how many solutions in the inteval 0<x<2pie to the following of (2xsin3x1)(cos2x+1)=0
anonymous
 one year ago
how many solutions in the inteval 0<x<2pie to the following of (2xsin3x1)(cos2x+1)=0

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hey, start by applying your `Zero Factor Property`,\[\large\rm 2x \sin3x1=0\qquad\qquad\qquad \cos2x+1=0\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Are you sure that first one has an \(\large\rm x\) and \(\large\rm sin3x\) in it? :o Hmm that's going to be tricky..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The second one shouldn't be too bad. Subtract 1 from each side,\[\large\rm \cos(\color{orangered}{2x})=1\]Recall that cosine is 1 at an angle of pi.\[\large\rm \color{orangered}{2x=\pi+2k \pi}\]Any number of full spins will get us back to that same point, so we add an amount of 2pi's on the end like that. Solving for x,\[\large\rm x=\frac{\pi}{2}+k \pi\]For k=0, we get \(\large\rm x=\frac{\pi}{2}\) For k=1, we get \(\large\rm x=\frac{3\pi}{2}\) Any other values of k will take us outside of our interval. So that's all the solutions we're getting from the second bracket.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hopefully that makes sense. Hmm I'm not sure what to do about the other bracket though :p

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh, it simply says "how many", not to actually find them... oh oh interesting.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Is this for calculus? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1cause maybe we could count the critical points.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your help and Yes it is for calculus. :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's see if any of these smart guys have an idea :d @ganeshie8 @Kainui @dan815
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