Task/26082833
--------------------
sin² x *cos⁻⁴x - 4 tg²x +3cos⁻² x -12 =0 ;
sin² x /cos⁴x - 4 tg²x +3*1/cos²x -12 =0 ;
tg² x* 1/cos²x -4tg²x +3*1/cos²x -12 = 0 ;
tg² x*( 1 +tg²x) - 4tg²x +3(1+tg²x) -12 =0 * * * 1+ tg²x =1/cos²x * * *
tg⁴x +tg² x - 4tg²x+3 +3tg²x -12 =0 ;
tg⁴x =9 ⇔
tg²x =3 ⇔(1-cos2x ) /(1+cos2x) =3 ⇔cos2x = -1/2 ;
2x = ±(π -π/3) +2πn , n∈ Z
x = ±π/3 +πn , n∈ Z .
ответ: ±π/3) +πn , n∈ Z .