Решить уравнение sin2x+4cos2x=1

sin2x+4cos2x=1;

2sinxcosx + 4cos^2x - 4sin^2x = 1;

2sinx/cosx + 4 - 4tg^2x = 1;

2tgx + 4 - 4tg^2x - 1 = 0;

-4tg^2x + 2tgx + 3 = 0;

4tg^2x - 2tgx - 3 = 0;

4t^2 - 2t - 3 = 0;

D = 4 + 4 * 4 * 3 = 52;

t = (2 + - 2sqrt13) / 8 = (1 + - sqrt13) / 4;

tgx = (1 + - sqrt13) / 4;

x = arctg ((1 + - sqrt13) / 4) + pin, n ∈ Z.