Trig and verifying Identities?

Question

Ok I'm not one of those lazy people who post they homework online and hope for the best. I'm using this as my last resort. I have been working on this for 1:30, and I have a test on it tomorrow. I have already done 30 other problems and I can't find the answer to these.

Find all solutions [0,2pi)

2(sin^2)2x + 5sin2x - 3 = 0

(2 (tan^2)x - 1)((tan^2)x +1) = 0

2 (sin^2)2x + 5 sin(2x) - 3 = 0

Verify:

[cot^2 x - tan^2 x)]/[(cot x + tan x)^2] = 2(cos^2) x - 1

Answer 1

1. Let u= Sin2x
Then 2u^2+5u-3=0
(u+3)(2u-1)=0
For u+3: sin 2x = -3 No solution
For 2u-1: 2 sin2x = 1 or sin 2x = 0.5
This happens at 2x=pi/6 and 5 pi/6 so x= pi/12 and 5pi/12. Since in the next cycle, 2x= 13 pi/6,
x=13 pi/12. Moreover 2x=17 pi/6, so x= 17pi/12.
Likewisex= 21pi/12.

2. Solve each factor separate.

3. Start with denominator:
cot x= cos x / sin x and tan x = sin x/cos x
Then (cot x+tanx)= [cos^2 x+sin^2 x]/ [sinx cos x] = 1/[sin x cos x], then, left hand side is
(sin^2 x cos^2) x (cot^2 x-tan^2 x)= cos^4 x - sin^4 x. This is the sum and diff of squares or
(cos^2 x + sin^2 x)(cos^2 - sin^2 x). First term is 1, so = cos^2 x - sin^2 x . Since sin^2 x=
1 - cos^2 x, term becomes 2 (cos^2 x )- 1