Question

how many solutions in the inteval 0 10 years ago

Answer 1

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

Answer 2

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..

Answer 3

Yes

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

Is this for calculus? :)

Answer 8

cause maybe we could count the critical points.

Answer 9

hmm

Answer 10

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

Answer 11

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