3 thằng đó lập thành csn thì ta có
$\cos^2 (x-\dfrac{7\pi}{12})=\cos (x-\dfrac{\pi}{3}) .\cos (x+\dfrac{\pi}{6})$
$\Leftrightarrow \dfrac{1+\cos (2x -\dfrac{7\pi}{6})}{2} =\dfrac{1}{2} \bigg [\cos (2x-\dfrac{\pi}{6}) +\cos \dfrac{\pi}{2} \bigg ]$
$\Leftrightarrow 1+\cos (2x -\dfrac{7\pi}{6})=\cos (2x -\dfrac{\pi}{6})$
$\Leftrightarrow 1-\cos (2x -\dfrac{\pi}{6})=\cos (2x -\dfrac{\pi}{6})$
$\Leftrightarrow \cos (2x -\dfrac{\pi}{6})=\dfrac{1}{2}$ quá dễ tự làm nốt đê