cos[(n+1)x]sin[(n+2)x]+cos[(n+1)x]cos[(n+2)x] = cos(x)

Let A = (n+1)x = nx+x

let B = (n+2)x = nx+2x

Then B-A = (nx+2x)-(nx+x) = nx+2x-nx-x = x

cos(A)sin(B)+cos(A)cos(B)=cos(B-A)


cos(A)sin(B)+cos(A)cos(B)=cos(B)cos(A)+sin(B)sin(A)


cos(A)sin(B) = sin(B)sin(A)

cos(A)sin(B) - sin(B)sin(A) = 0

sin(B)[cos(A)-sin(A)} = 0


sin(B) = 0;             cos(A)-sin(A) = 0
                        -sin(A) = -cos(A)
B = ;               sin(A) = cos(A)
where P is any integer  
(n+2)x = ;          tan(A) = 1 
x =                A =  
                        A = 
                        where Q is any integer
                        n(x+1) = 
                        x+1 = 
                        x =     
Edwin