## anonymous one year ago how many solutions in the inteval 0<x<2pie to the following of (2xsin3x-1)(cos2x+1)=0

1. zepdrix

Hey, start by applying your Zero Factor Property,$\large\rm 2x \sin3x-1=0\qquad\qquad\qquad \cos2x+1=0$

2. 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..

3. anonymous

Yes

4. 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.

5. zepdrix

Hopefully that makes sense. Hmm I'm not sure what to do about the other bracket though :p

6. zepdrix

Oh, it simply says "how many", not to actually find them... oh oh interesting.

7. zepdrix

Is this for calculus? :)

8. zepdrix

cause maybe we could count the critical points.

9. zepdrix

hmm

10. anonymous

Thanks for your help and Yes it is for calculus. :)

11. zepdrix

Let's see if any of these smart guys have an idea :d @ganeshie8 @Kainui @dan815