A) cos(Pi/2+x)=-sin(Pi/6);
cos(Pi/2+x)=-1/2;
Pi/2+x=+-arccos(-1/2)+2Pi*n, n є Z;
Pi/2+x=+-2Pi/3+2Pi*n, n є Z;
x=+-2Pi/3-Pi/2+2Pi*n, n є Z;
б) cos^2x-sin^2x=1+4cosx;
cos^2x-1+cos^2x=1+4cosx;
2cos^2x-2-4cosx=0;| :2;
cos^2x-2cosx-1=0;
cosx=t;
t^2-2t-1=0;
D=4+4=8;
t1=(2+sqrt8)/2=1+sqrt2;
t2=1-sqrt2;
t1>1;
cosx=1-sqrt2;
x=+-arccos(1-sqrt2)+2Pi*n, n є Z
в) 3tgx=-sqrt3;
tgx=-sqrt3/3;
x=arctg(-sqrt3/3)+Pi*n, n є Z;
x=5Pi/6+Pi*n, n є Z