Dividing by sin(x) in inequalities and factoring cubic with imaginary roots

[epkid08]

I have this review packet and I have a couple questions.

Homework Statement

Solve the fallowing by factoring and making appropriate sign charts.

$$2sin^2(x)>sin(x), 0\leqX\leq2\pi$$

Homework Equations

The Attempt at a Solution

Well my main question is, can I divide both sides by sin(x)? In doing so, I got $$2sin(x)\geq1$$, and $$2sinx\leq1$$ because sin(x) can be both positive and negative. After solving I got $$x\leq\pi/6 \leq x$$, which seems pointless.

Okay tex isn't working, the last sentence should say, "x is less than or equal to pi divided by 6 which is less than or equal to x.

Homework Statement

Use p/q method to factor the polynomial, then solve for P(x)=0.

P(x) = x^3 - 6x^2 + 3x - 10

Homework Equations

The Attempt at a Solution

The p/q method doesn't seem to work here because there are two imaginary roots. How can I factor this?
I'll probably end up posting more. Thanks for the help

 

[yeongil]

Homework Statement

Solve the fallowing by factoring and making appropriate sign charts.
$$2sin^2(x)>sin(x), 0\leqX\leq2\pi$$

Homework Equations

The Attempt at a Solution

Well my main question is, can I divide both sides by sin(x)? In doing so, I got $$2sin(x)\geq1$$, and $$2sinx\leq1$$ because sin(x) can be both positive and negative. After solving I got $$x\leq\pi/6 \leq x$$, which seems pointless.

Okay tex isn't working, the last sentence should say, "x is less than or equal to pi divided by 6 which is less than or equal to x.

I think that when you divide by sin(x) you lose a root (or roots). Instead, subtract both sides by sin(x), factor the left side, and go on from there.


01

 

[HallsofIvy]

The problem said "by factoring and making appropriate sign charts."

Look at 2sin²(x)- sin(x)= sin(x)(2sin(x)- 1)> 0. The "appropriate sign charts" refers to the fact that a product is positive as long as the factors have the same sign. What are the signs of sin(x) and 2sin(x)- 1 over various intervals? Where are they the same?

As for the last problem, if two of the roots are complex, then the third must be real: your polynomial can be factored into linear and quadratic factors. I suspect that the "p/q method" is what I would call the "rational root theorem": if a polynomial equation, with integer coefficients, has a rational root, p/q, the p must divide the constant term and q must divide the leading coefficient. Here the leading coefficient is 1 and the constant term is 10. That tells you that there are only 4 possible rational roots. Plug each of them into the equation and see if one satisfies it.

 

[epkid08]

As for the last problem, if two of the roots are complex, then the third must be real: your polynomial can be factored into linear and quadratic factors. I suspect that the "p/q method" is what I would call the "rational root theorem": if a polynomial equation, with integer coefficients, has a rational root, p/q, the p must divide the constant term and q must divide the leading coefficient. Here the leading coefficient is 1 and the constant term is 10. That tells you that there are only 4 possible rational roots. Plug each of them into the equation and see if one satisfies it.

I am sure I checked all of the possible numbers at least twice, I guess I'll check again.

Edit: It is possible that the only real root left is irrational, so there is no way to factor it? (how would I find the root?)

 

[HallsofIvy]

Yes, it is possible that the only real root of a polynomial is irrational. In that case, while it, strictly speaking, can be factored, it would be extremely difficult to find the real root and so factor. But are you sure you have copied the problem correctly?
x³- 6x²+ 3x+ 10 has a simple integer root.