1=cosxsin2x-sinxcos2x/sin2x

Question

solve the function below for xe[0, 2pi]

1=cosxsin2x-sinxcos2x/sin2x

Answer 1

1=(cos(x)sin(2x)-sin(x)cos(2x))/sin(2x).

sin(A-B)≡sinAcosB-cosAsinB (trig identity).

Let A=2x and B=x, sin(2x-x)=sin(x)=sin(2x)cos(x)-cos(2x)sin(x), that is:

cos(x)sin(2x)-sin(x)cos(2x)=sin(x).

Therefore 1=sin(x)/sin(2x)=sin(x)/(2sin(x)cos(x))=1/(2cos(x)), provided x≠nπ, where n is an integer, including zero.

So, 2cos(x)=1, cos(x)=½, x=⅓π or 2π-⅓π=5π/3. (x=60° or 300°)