Ответ:
b)2cos^2x+sin4x=1
2cos^2x+sin4x-1=0
2cos^2x-1 = cos2x — формула двойного угла
cos2x+sin4x=0
cos2x+2*sin2x*cos2x=0
cos2x(1+2sin2x)=0
cos2x=0 ->2x=pi/2+pi*k ->x=pi/4+pi*k/2
1+2sin2x=0 ->sin2x= -1/2 -> 2x=(-1)^(n+1) *pi/6 +pi*n ->x=((-1)^(n+1) *pi)/12 +pi*n/2
Объяснение:
A)sinπ/33*cosπ/33*cos2π/22*cos4π/33*cos8π/33*cos16π/33/sinπ/33=
=sin2π/33*cos2π/22*cos4π/33*cos8π/33*cos16π/33/(2sinπ/33)=
=sin4π/33*cos4π/33*cos8π/33*cos16π/33/(4sinπ/33)=
=sin8π/33*cos8π/33*cos16π/33/(8sinπ/33)=sin16π/33*cos16π/33/(16sinπ/33)=
=sin32π/33/(32sinπ/33)=sin(π-π/33)/32sinπ/33=(sinπ/33)/32sinπ/33=1/32
b)cosπ/7cos4π/7cos(π-2π/7)=-cosπ/7cos2π/7cos4π/7=
=(-sinπ/7cosπ/7cos2π/7cos4π/7)/(sinπ/7)=(-sin2π/7cos2π/7cos4π/7)/(2sinπ/7)=
=(-sin4π/7cos4π/7)/(4sinπ/7)=(-sin8π/7)/(8sinπ/7)=(-sinπ/7)/(8sinπ/7)=-1/8
2a-b=7 - b=7-2a b=2a-7 3a-2(2a-7)=14 3a-4a+14=14 -a=0 a=0
(-0,5a¹⁵b⁴)² · (-4/7ab⁷) = 0,25а³⁰b⁸ · (-4/7ab⁷) = 1/4a³⁰b⁸ · (-4/7ab⁷<span>) = -1/7a</span>³¹b¹⁵