Trigonometry
posted by Anonymous .
cot^2(3x)sin(2x)sin(2x)=0
2sin^2(5x)+sin(5x)1=0

For the first, take out a common factor of sin(2x) to get
sin(2x)[cot^2(3x)  1] = 0
now you have a difference of squares for
sin(2x)(cot(3x  1)(cot(3x) + 1) = 0
Can you take it from there?
I will do the second one:
Your equation is a quadratic which factors.
(2sin(5x)  1)(sin(5x) + 1) = 0
sin(5x) = 1/2 or sin(5x) = 1
5x = 30º or 150º or 5x = 270º
x = 6º , 30º , 54º
I don't know what your domain was, but the period of sin(5x) is 360/5º or 72º
so by adding or subtracting multiples of 72 to each of my answers will yield as many answers as you want.
general solutions in radians :
pi/30 + k(2pi/5)
pi/6 + k(2pi/5)
3pi/10 + k(2pi/5) , where k is an integer. 
Thank you! Would you mind explaining to me the k(2pi/5) thing? How do you know when to use that and how do you know when to use just k(pi/5) without the 2? Is the period always going to be the number in front of x (like in this case it was 5 because of 5x)?