Verified answer

sin²x+cos²x=1

sin(2x)=2sinx*cosx

sin⁴x+cos⁴x=sin²(2x)-1/2

(sin²x+cos²x)²-2sinx*cosx=sin²(2x)-1/2;

sin²(2x)-1-sin(2x)-1/2=0

sin(2x)=t;  ItI≤1;

t²-t-3/2=0

D=1+6=7

t=(1±√7)/2

t=(1+√7)/2∈∅

t=(1-√7)/2

sin(2x)=(1-√7)/2

2x=(-1)ⁿarcsin((1-√7)/2)+πn; n∈Z;

x=0.5*(-1)ⁿarcsin((1-√7)/2)+πn/2; n∈Z;

6) cosx+3sinx=2

cosx=cos²(x/2)-sin²(x/2)

sin(x)=2sin(x/2)*cos(x/2)

cos²(x/2)-sin²(x/2)+6sin(x/2)*cos(x/2)=2*(cos²(x/2)+sin²(x/2))

-cos²(x/2)-3sin²(x/2)+6sin(x/2)*cos(x/2)=0

разделим обе части на -cos²(x/2)≠0

1+3tg²(x/2)-6tg(x/2)=0

tg(x/2)=t

3t²-6t+1=0

D=36-12=24

t=(6+√24)/6=(1+√6/3)

t=(6-√24)/6=((1-√6/3)

tg(x/2)=(1+√6/3)

x/2=arctg(1+√6/3)+πn; n∈Z

x=2arctg(1+√6/3)+2πn; n∈Z

tg(x/2)=(1-√6/3)

x/2=arctg(1-√6/3)+πк; к∈Z

x=2arctg(1-√6/3)+2πк;к∈Z