1 = sqrt(3)tan(6 theta)

theta = Pi/36

a million + tan(?)sin(?) = 0 a million + sin²(?)/cos(?) = 0 [cos(?) + sin²(?)]/cos(?) = 0 cos(?) + sin²(?) = 0 a million + cos(?) - cos²(?) = 0 cos²(?) - cos(?) - a million = 0 cos(?) = [a million ± ?(5)]/2 cos(?) = [a million - ?(5)]/2 ... Reject the useful answer. this is greater than one. ? = cos?¹{[a million - ?(5)]/2} i might desire to provide you a decimal approximation for that, yet as for an exact answer, this is so a ways as i will carry it. Edit: I see now which you wrote 2 different issues. Above I solved the 1st one, the only without branch sign. right this is the 2nd. a million + tan(?)/sin(?) = 0, sin(?) ? 0 a million + a million/cos(?) = 0 cos(?) + a million = 0 cos(?) = -a million ? = (2k + a million)?, for some integer ok in spite of the shown fact that, for all of those ideas, sin(?) may well be equivalent to 0, so as that they might desire to all be rejected. hence, the equation has no answer on any era.

1 = sqrt(3) * tan(6t)

1 / sqrt(3) = tan(6t)

sqrt(3) / 3 = tan(6t)

6t = pi/3 + pi * k

t = pi/18 + pi * k / 6

k is an integer

t = pi * (1/18 + 0/6) = pi * (1/18)

t = pi * (1/18 + 1/6) = pi * (4/18) = pi * (2/9)

t = pi * (1/18 + 2/6) = pi * (7/18)

t = pi / 18 , 2 * pi / 9 , 7 * pi / 18