can anyone help me solve these equations?

Question

im have to solve these (w/ answers between 0 and 2pi) using identities, and i know them but after applying them i keep ending up stuck with an equation i cant solve...

these are the questions:
a) 2 csc^2x = 3 cot^2x - 1
b) sin ^2x + sin x - 1 = 0
c) 3 cosxcotx + 7 = 5 cscx
d) 2 cos^2x - cos x = 2 - secx

can anyone help with any of them?

Answer 1

a)2csc^2(2x) = 3cot^2x -1
[csc^2 = 1 + cot^2]
2(1 + cot^2(2x)) = 3cot^2x -1
2 + 2cot^2(2x) = 3cot^2x - 1
3 = cot^2x
cot2x = 1/√3
2x = 60 x = 30

b)sin^2x + sinx - 1 = 0
Let u = sinx
u^2 + u -1 = 0
u = (-1 +/- √(1 + 4))/2
sinx = (-1 +/- √5)/2
x = 38.1727degrees, the other answer is extraneous

c)3cosxcotx + 7 = 5cscx
3cosx(cosx/sinx) + 7 = 5/sinx
3cos^2x/sinx + 7 = 5/sinx
3(1 - sin^2x)/sinx + 7 = 5/sinx
3/sinx - 3sinx + 7 = 5/sinx
-3sinx + 7 - 2/sinx = 0
3sin^2x - 7sinx + 2 = 0
(3sinx - 1)(sinx - 2) = 0
3sinx = 1
sinx = 1/3
x = 19.47122degrees, the other factor is extraneous

d) 2cos^x - cosx = 2 - secx
2cos^2x - cosx = 2 - 1/cosx
2cos^3x - cos^2x = 2cosx - 1
2cos^3x - 2cosx - cos^2x + 1 = 0
2cosx(cos^2x - 1) + (-1)(cos^2x - 1) = 0
(2cosx - 1)(cos^2x - 1) = 0
2cosx = 1
cosx = 1/2
x = 60degrees
cos^2 x = 1
cosx = +/- 1
x = 0, 180