Solving a Trig equation using a trig identity

Please could someone just check this is right

The Question: 3 sin² x = 1 + 5 cos x

Using the identity sin² x = 1 - cos² x
3(1 - cos² x) = 1 + 5 cos x
3 – 3 cos² x = 1 + 5 cos x
1 + 5 cos x = 3 – 3 cos² x
3 cos² x + 5 cos x + 1 = 3
3 cos² x + 5 cos x – 2 = 0
3c² + 5c – 2 = 0

Using the quadratic formula
A = 3, b = 5 and c = -2

-5 plus and minus √(5² - 4x3x-2)
2x3

-5 plus and minus √(25 - - 24)
6

-5 plus and minus √49
6

-5 + 7
6
2/6 = 1/3

-5 -7
6
-12/6 = -2

As I am looking for answers in the range of 0 and 360 degrees, I am just going to concentrate on the positive answer, which is 1/3

So (Cos to the power of minus one) 1/3 = 70.5 degrees
70.5 + 180 = 250.5 degrees

Are these the correct answers???

Please could someone just check this is right

The Question: 3 sin² x = 1 + 5 cos x

Using the identity sin² x = 1 - cos² x
3(1 - cos² x) = 1 + 5 cos x
3 – 3 cos² x = 1 + 5 cos x
1 + 5 cos x = 3 – 3 cos² x
3 cos² x + 5 cos x + 1 = 3
3 cos² x + 5 cos x – 2 = 0
3c² + 5c – 2 = 0

(3c-1)(c+2) = 0
= 3c² - c + 6c - 2
= 3c² + 5c - 2.

So c = 1/3 or -2.

Thank you, I was trying to do it this way first, but didn't think to put the 2 in a different bracket as the 3c to get the 6c - im a dumbass at times haha