Solve for x(basic question for fresher)

[Proleague]

Homework Statement

I totally forgot my basic pre-calculus stuff.
So, please help me out to do this.

Homework Equations

Solve for x
1. (x-4)(x+5)>0

2. sin2x=sinx, 0<=x<=2x

The Attempt at a Solution

I tried number 1 it said x=4, x=-5 but it should be greater than 0.
I just confused only this part.

And, number 2 I don't know what the hell is this. T_T

 

[Mark44]

For 1, if a*b > 0, what can you say about a and b?
BTW, x = 4 and y = -5 are NOT solutions of the inequality.
For 2, do you remember any of the double angle trig identities?

 

[Proleague]

I forgot almost things. so, For 1 I just answered x>4 and x>-5 is it correct?

and, I tried number2 using sin 2x = 2sin(x)cos(x)
so, I just got 2cox=1 but, I don't know next step. Please help me out.

 

[ideasrule]

I forgot almost things. so, For 1 I just answered x>4 and x>-5 is it correct?

and, I tried number2 using sin 2x = 2sin(x)cos(x)
so, I just got 2cox=1 but, I don't know next step. Please help me out.

Think about the first problem this way:

If x is really negative, say negative one million, then x-4 and x+5 will both be negative and (x-4)(x+5) would both be positive. As we increase x, x-4 and x+5 will remain negative until x becomes -5, in which case x+5 becomes zero. Increase x a tiny bit further and x+5 would be positive while x-4 would still be negative, so (x-4)(x+5) < 0. Increase x past x=4 and x-4 would be positive too. Increase it even more and (x-4)(x+5) would stay above 0 because both factors stay positive. So the intervals that where (x-4)(x+5) > 0 are...

As for the second problem, I supposed you divided both sides of sin x = 2sin(x)cos(x) by sin(x). Don't do that; you don't know that sin(x) isn't 0, and if it is, you'll be dividing by 0. Instead, rearrange the equation and factor it so that one side is 0. That way, you could say "if either factor 1 is 0 or factor 2 is 0, the whole thing would be 0. Factor 1 is zero when...factor 2 is 0 when..."