Questions about trig identities

Given that 2cos²x - sin²x = 1 Show that cos²x = 2sin²x

Given that 5sec²x + 3tan²x = 9 find the possible values of sinx
(I tried using 1 + tan²x = sec²x)

Thanks for any help given :)

2cos^2x - sin^2x = 1
2cos^2x - sin^2x = cos^2x + sin^2x
cos^2x = 2sin^2x

5sec^2x + 3tan^2x = 9
5sec^2x + 3(sec^2x - 1) = 9
5sec^2x + 3sec^2x - 3 = 9
8sec^2x = 12
sec^2x = 12/8 = 3/2
cos^2x = 2/3
sinx = +/-(1 - cos^2x)^(1/2)
sinx = +/-(1 - 2/3)^(1/2) = +/-(1/3)^(1/2)

These are not trig identities. It is not true that 2cos²x - sin²x = 1. Neither is it true that 5sec²x + 3tan²x = 9. If you are looking to find specific values of x for which these equations are satisfied then you should think about rewording your question.