Help please! im stuck with a question, maths homework

I cant do these two :(
6sec^2(2x) + 5tan2x = 12

and

3sin^2(3x)-7cos3x-5=0

please help!
on the forst one i have started and thin i may have reached 6=10sinxcos^3x-10sin^3xcosx
but this could be wrong.. and what now??

Hey there! It's gonna be a long one, so get ready.

Okay here's what we're going to do. Also, I'll assume that you're allowed to use calculators. :D

1. convert sec2 2x into 1 + tan2 2x.
We now get:
6(1 + tan2 2x) + 5tan 2x = 12
6 + tan2 2x + 5tan 2x = 12
tan2 2x + 5tan 2x = 6
Let u = tan 2x and we get:
u^2 + 5u - 6 = 0 --> QUADRATIC EQUATION!
(u + 6)(u-1) = 0
Solve for u:
u = -6 OR u = 1.
tan 2x = -6 (A) OR tan 2x = 1 (B)
(A):
2x = arctan -6
x = (arctan -6 )/2
(B):
2x = arctan 1 (obviously, arctan 1 is pi/4 because that's the angle when sin and cos are equal!)
x = pi/8
ANSWERS: -0.7028 and pi/8

2. convert sin2 3x into 1 - cos2 3x
It's just the same as the first one. Substitute u into cos 3x and you'll get another quadratic equation.
We now get:
3u^2 + 7u + 2 = 0
(3u + 1)(u + 2) = 0
Solve for u:
u = -1/3 OR u = -2.
WARNING: u = -2 is IMPOSSIBLE. That's because cosine of whatever will never be larger(or lesser) than 1(or -1). Same goes with sine.
THE ONLY ANSWER:
x = (1/3)*arccos (1/3) or 0.63688

WTH, Now I can't even read my own answer!

But the program that I did was just some random shiz. It's just a glorified calculator. (I can't find my calc so I just did it on my computer). Program was written using the Python programming language :D

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