Ответ на первую картинку 2,4
на вторую хз
Sina-sinb=2sin[(a-b)/2]*cos[(a+b)/2]
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sin2x-sin(π/2-2x)=√2sin3x
2sin(2x-π/4)*cosπ/4=√2*sin3x
2*√2/2*sin(2x-π/4)=√2*sin3x
√2*sin(2x-π/4)=√2*sin3x
sin(2x-π/4)=sin3x
sin(2x-π/4)-sin3x=0
2sin(-x/2-π/8)*cos5x/2-π/8)=0
-2sin(x/2+π/8)*cos(5x/2-π/8)=0
sin(x/2+π/8)=0⇒x/2+π/8=πk⇒x/2=-π/8+πk⇒x=-π/4+2πk,k∈z
cos(5x/2-π/8)=0⇒5x/2-π/8=π/2+πk⇒5x/2=5π/8+πk⇒x=π/4+2πk/5,k∈z