Question

A) 1-2cos^2x=sin(p-x) b)9p/2;13p/2

Answer 1

1-2(1-sin^2(x))=sin(x);
-1+2sin^2(x))=sin(x);  
2sin^2(x))-sin(x)-1=0;  
замена:  y=sin(x)   =>  2y^2-y-1=0   =>  y1= 1;  y2= -1/2     =>  
1)  sin(x) = 1    =>  x=p/2+pn;  
2)  sin(x) = -1/2   =>  x= - p/6 + 2pn