Решить sin^2x+sin^2(2x)+sin^2(3x)+sin^2(4x)=2 - вопрос №559563722223242 от hikka137 17.04.2021 07:08

Question

Алгебра: Решить sin^2x+sin^2(2x)+sin^2(3x)+sin^2(4x)=2

hikka137 · 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