anonymous
  • anonymous
how many solutions in the inteval 0
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
Hey, start by applying your `Zero Factor Property`,\[\large\rm 2x \sin3x-1=0\qquad\qquad\qquad \cos2x+1=0\]
zepdrix
  • zepdrix
Are you sure that first one has an \(\large\rm x\) and \(\large\rm sin3x\) in it? :o Hmm that's going to be tricky..
anonymous
  • anonymous
Yes

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zepdrix
  • zepdrix
The 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
  • zepdrix
Hopefully that makes sense. Hmm I'm not sure what to do about the other bracket though :p
zepdrix
  • zepdrix
Oh, it simply says "how many", not to actually find them... oh oh interesting.
zepdrix
  • zepdrix
Is this for calculus? :)
zepdrix
  • zepdrix
cause maybe we could count the critical points.
zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
Thanks for your help and Yes it is for calculus. :)
zepdrix
  • zepdrix
Let's see if any of these smart guys have an idea :d @ganeshie8 @Kainui @dan815

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