4tg^2 (x) - 3/(sin (3pi/2+x)) +3 = 0

sin(3π/2 + x) = -cosx (как получилось - во вложениях)
tg²x = sin²x/cos²x
$\frac{4sin^2x}{cos^2x} + \frac{3}{cosx} + 3 = 0 \\ * na cos^2x \\ 4sin^2x + 3cosx + 3cos^2x = 0 \\ 4(1-cos^2x) + 3cosx + 3cos^2x = 0 \\ 4 - 4cos^2x + 3cosx + 3cos^2x = 0 \\ -cos^2x + 3cosx + 4 = 0\\ cos^2x - 3cosx - 4 = 0 \\ cosx = t \\ t^2 - 3t - 4 = 0 \\ D = 9-4*1*(-4) = 9 + 16 = 25 =\ \textgreater \ 5 \\ t1 = (3+5)/2 = 4 \\ t2 = (3-5)/2 = -1 \\ cosx = 4 net korney \\ cosx = -1 \\ x = p + 2pik (vo vlozheniyax, esli nuzhno) \\$