find all of the solutions of the equation in the interval (0, 2pi) work and steps please?

Question

work and all steps please
1. sec squared x - 1 = 0
2. 2 sin ^2x = 2 + cos x
3. secx cscx= 2 csc x

Answer 1

1. On the left-hand side, this equals Tan^2 x.
So Tan^2 x = 0 => Tan x =0. This happens at 0, pi and 2 pi.
2. If you divide by 2, you get
sin^2 x = 1 + cos x/2
I believe this only happens at pi/2
3. Since you have csc x on both sides, you can divide to get sec x = 2. Based on it being reciprocal cosine, sec x=2 happens when cosine x=1/2. This occurs at pi/3 and 5 pi/3

Answer 2

1. 2cos^2x - cosx = 0 . . . factor cos x cos x (2 cos x -1) = 0 cos x = 0. . . . . x = 0º 2 cos x -1 = 0 . . . cos x = 1/2 . . . x = 60º . . & . . x = 330º 2. 3tanx - cotx = 0 3 sin x / cos x = cos x / sin x cos² x = 3 sin x 1 - sin² x = 3 sin x sin² x + 3 sin x - 1 = 0 sin x = 0.30278 .. x = 17.6248º. . . and . . x = 180-x =162.375º 3. cos² x - 5 cosx - 1 = 0 in the interval [0,2pi) cos x = -0.19258. . . . . . by quadratic equation x = 78.8966º . . . and . . . x = 180-x = 101.103º