- #1
epkid08
- 264
- 1
I have this review packet and I have a couple questions.
Solve the fallowing by factoring and making appropriate sign charts.
[tex]2sin^2(x)>sin(x), 0\leqX\leq2\pi[/tex]
Well my main question is, can I divide both sides by sin(x)? In doing so, I got [tex]2sin(x)\geq1[/tex], and [tex]2sinx\leq1[/tex] because sin(x) can be both positive and negative. After solving I got [tex]x\leq\pi/6 \leq x[/tex], 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.
Use p/q method to factor the polynomial, then solve for P(x)=0.
P(x) = x^3 - 6x^2 + 3x - 10
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
Homework Statement
Solve the fallowing by factoring and making appropriate sign charts.
[tex]2sin^2(x)>sin(x), 0\leqX\leq2\pi[/tex]
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 [tex]2sin(x)\geq1[/tex], and [tex]2sinx\leq1[/tex] because sin(x) can be both positive and negative. After solving I got [tex]x\leq\pi/6 \leq x[/tex], 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