Chào bạn!
nếu mình ko nhầm thì đề bài như này:
$\dfrac{sin^3\dfrac{x}{2}-cos^3\dfrac{x}{2}}{2+sinx}=\dfrac{1}{3}cox$
$\Leftrightarrow3(sin^3\dfrac{x}{2}-cos^3\dfrac{x}{2})=cox(2+sinx)$
$\Leftrightarrow3(sin\dfrac{x}{2}-cos\dfrac{x}{2})(sin^2\dfrac{x}{2}+sin\dfrac{x}{2}cos\dfrac{x}{2}+cos^2\dfrac{x}{2})=(cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2})(2+2sin\dfrac{x}{2}cos\dfrac{x}{2})$
$\Leftrightarrow(sin\dfrac{x}{2}-cos\dfrac{x}{2})[3(sin\dfrac{x}{2}+cos\dfrac{x}{2})^2-3sin\dfrac{x}{2}cos\dfrac{x}{2}-(cos\dfrac{x}{2}+sin\dfrac{x}{2})(2+2sin\dfrac{x}{2}cos\dfrac{x}{2}]=0$