\(cos2\left(x+\dfrac{\pi}{3}\right)=cos2\left[\dfrac{\pi}{2}-\left(\dfrac{\pi}{6}-x\right)\right]=cos\left[\pi-2\left(\dfrac{\pi}{6}-x\right)\right]\)
\(=-cos2\left(\dfrac{\pi}{6}-x\right)=1-2cos^2\left(\dfrac{\pi}{6}-x\right)=1-2t^2\)
Phương trình đã cho trở thành:
\(1-2t^2+4t=\dfrac{5}{2}\Leftrightarrow4t^2-8t+3=0\)