Question

6cos^2 (4x) + ctg * 4x - 1 = 0

Answer 1

Ответ:

6cos^2(4x)+ctg*4x-1=0,cos4x=t t6^2+t-1=0, D=1+24=25t=-1+-5/12=-1/2 и 1/3

1)cos4x=-1/3

4x=arccos(-1/2)+2pn

4x=pi-pi/3+2pn

4x=2pi3+2pn

x=pi/6+pn/2