Answer:
Hello,
Step-by-step explanation:
I suppose the question is to solve the equation.
[tex]Remember:\\\\sin(a)-sin(b)=2cos(\dfrac{a+b}{2} )*sin(\dfrac{a-b}{2} )\\cos(a)-cos(b)=2sin(\dfrac{a+b}{2} )*cos(\dfrac{b-a}{2} )\\\\\\sin(2a)+cos(4a)=cos(2a)+sin(4a)\\\\\Longleftrightarrow sin(4a)-sin(2a)=cos(4a)-cos(2a)\\\\\Longleftrightarrow 2cos(\dfrac{4a+2a}{2})*sin(\dfrac{4a-2a}{2})=2cos(\dfrac{4a+2a}{2})*sin(\dfrac{2a-4a}{2})\\\\\Longleftrightarrow cos(3a)*sin(a)=cos(3a)*(-sin(a))\\\\\Longleftrightarrow 2cos(3a)*sin(a)=0\\\\[/tex]
[tex]sin (a)=0\ or\ cos(3a)=0\\\\\boxed{k\in\ \mathbb{Z}\ : a=k*\pi\ or\ a=\dfrac{(2k+1)*\pi}{2} }\\[/tex]
