Question

Помогите пожалуста))) sin7x+sin3x=3cos2x

Answer 1

Sin7x+sin3x=3cos2x;⇒2sin(7x+3x)/2·cos(7x-3x)/2=3cos2x;⇒
2sin5x·cos2x-3cos2x=0;⇒cos2x(2sin5x-3)=0;
cos2x=0;⇒2x=π/2+kπ;k∈Z;⇒x=π/4+kπ/2;k∈Z;
2sin5x-3=0;⇒sin5x=3/2;3/2>1;-решений нет,т.к 
-1≤sinx≤1;