A couple of Pre-Calculus problems.

[Kerpew]

Okay, here's the scoop. I took a test on Wednesday and got it back on Friday. Our teacher is trying a new experiment, the super special awesome "test corrections". The idea is, we go and find what we missed, type up in detail what our mistakes were, then redo the problem showing all of our work. Our reward and motivation is gaining up to half of the point of what we missed. We have to do all of the problems perfectly, however, and it's an all-or-nothing assignment. I was able to figure out most of the things I missed, but two problems bug me now. And with that very long introduction to my problem aside, here's what the problems are:


For problems 4-5, find all solutions to the equation on the domain [0, 2?).

5. tanXsin²X = 2tanX

For this problem, I just didn't know where to begin at all, and I can't figure it out at all.


For problems 6-9, find all solutions.

9. 2cosX - cotX = 0

For this one, I can get really close to the answer. But I think I created too much work for myself and there may be an easier way.

I did this one like this:

I know from the Cosine I can get the answers:
?/2 + 2?n and 3?/2 + 2?n.

But I can't figure out how to get the answers with the Cosecant. ^^; Maybe I just did it all wrong.

If anyone could help, thanks a bunch!

 

[Deleted member 4993]

I know from the Cosine I can get the answers:
?/2 + 2?n and 3?/2 + 2?n............remember the restriction on domain of solutions

csec(x) = 2

sin(x) = 1/2

\(\displaystyle x = \frac{\pi}{6}\,\, and \, \frac{5\cdot \pi}{6}\)

4) tanX * sin[sup:2x9bcdfk]2[/sup:2x9bcdfk]X = 2tanX

tanX*(sin[sup:2x9bcdfk]2[/sup:2x9bcdfk]X - 2) = 0

Now follow the process of the other problem.

 

[pka]

#4 can be tricky. Look at the graphs. Clearly the two are equal if x=0.
Where else would equality happen?

 

[Dr. Flim-Flam]

tan(x)[sin(x)]^2 = 2tan(x), Domain 0=<x<2Pi

tan(x)[sin(x)]^2 - 2tan(x) = 0, tan(x)[(sin(x))^2-2] = 0

tan(x) = 0, x = 0, Pi, [sin(x)]^2 = 2, |sin(x)| = sqrt(2), however |sin(x)| <= 1, sqrt(2) =1.414... No can be hence

x =0 and x = Pi in the domain 0=< x < Pi

Next one: 2cos(x) - cot(x) = 0, domain All reals

2cos(x) - cos(x)/sin(x) = 0, [2cos(x)sin(x) - cos(x)]/sin(x) = 0, cos(x)[2sin(x) - 1]/sin(x) = 0.

Ergo cos(x) = 0, x = Pi/2 +kPi, k an integer and sin(x) = 1/2, x = Pi/6 + 2kPi and x = 5Pi/6 + 2kPi, k an integer.

Also note sin(x) does not = 0.