Решить ! sin^2 x +sin^2 2x + sin^2 3x + sin^2 4x =2 - вопрос №559949222223242 от rmaro82 15.05.2022 11:53

Question

Алгебра: Решить ! sin^2 x +sin^2 2x + sin^2 3x + sin^2 4x =2 кто нибудь

rmaro82 · Answer

(1-cos2x)/2+(1-cos4x)/2+(1-cos6x)/2+(1-cos8x)/2=2
1-cos2x+1-cos4x+1-cos6x+1-cos8x=4
(cos2x+cos8x)+(cos4x+cos8x)=0
2cos5xcos3x+2cos5xcos2x=0
2cos5x(cos3x+cos2x)=0
2cos5x*2cos(5x/2)cos(x/2)=0
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5,n∈z
cos(3x/2)=0⇒3x/2=π/2+πk⇒x=π/3+2πk/3,k∈z
cos(x/2)=0⇒x/2=π/2+πm⇒x=π+2πm,m∈z