1
5-5sinx=2cos²x
5-5sinx-2+2sin²x=0
sinx=a
2a²-5a+3=0
D=25-24=1
a1=(5-1)/4=1⇒sinx=1⇒x=π/2+2πn,n∈z
a2=(5+1)/4=1,5⇒sinx=1,5>1 нет решения
2
[(cosx+sinx)²]²+[(cosx-sinx)²]²=
=(cos²x+2cosxsinx+sin²x)²+(cos²x+2cosxsinx+sin²x)²=(1+sin2x)²+(1-sin2x)²=
=1+2sin2x+sin²2x+1-2sin2x+sin²2x=2+2sin²2x=2+2(1-cos4x)/2=
=2+1-cos4x=3-cos4x
3
arccos[cos(2arcctg(√2-1))]=2arcctg(√2-1)