sin^2 x/cos^2 x = 3/(2cosx)
2 cosx sin^2 x =3 cos^2 x
2cosx(1-cos^2 x) = 3cos^2 x
2cosx - 2cos^3 x - 3cos^2x = 0
-cosx(2cos2x + 3cosx - 2) = 0
cosx(cosx + 2)(2cosx - 1) = 0
cosx = 0 OR cosx = -2, not possible OR cosx= 1/2
x = 90° or 270° or 60° or 300°
in radians:
x = π/2, 3π/2, π/6 , 5π/6
tan^2(x)=-3/2sec(x)
2 answers
Or, reading more carefully, and noticing the "-" sign,
sin^2 x/cos^2 x = -3/(2cosx)
2 cosx sin^2 x = -3 cos^2 x
2cosx(1-cos^2 x) = -3cos^2 x
2cosx - 2cos^3 x + 3cos^2x = 0
-cosx(2cos^2x - 3cosx - 2) = 0
cosx(cosx - 2)(2cosx + 1) = 0
cosx = 0 OR cosx = 2, not possible OR cosx= -1/2
x = 90° or 270° or 120° or 240°
in radians:
x = π/2, 3π/2, 2π/3 , 4π/3
sin^2 x/cos^2 x = -3/(2cosx)
2 cosx sin^2 x = -3 cos^2 x
2cosx(1-cos^2 x) = -3cos^2 x
2cosx - 2cos^3 x + 3cos^2x = 0
-cosx(2cos^2x - 3cosx - 2) = 0
cosx(cosx - 2)(2cosx + 1) = 0
cosx = 0 OR cosx = 2, not possible OR cosx= -1/2
x = 90° or 270° or 120° or 240°
in radians:
x = π/2, 3π/2, 2π/3 , 4π/3