Môn Toán Lớp 11: giải pt a, cos2x – 2cosx.cos(x – π/3) +1 = 0

cos 2x-2cos x cos (x-\pi/3)+1=0

<=>2cos^2 x-1-2cos x cos (x-\pi/3)+1=0

<=>2cos^2 x-2cos x cos (x-\pi/3)=0

<=>2cos x[cos x-cos (x-\pi/3)]=0

<=> $\left[\begin{matrix} cos x=0\\ cos x=cos (x-\dfrac{\pi}{3})\end{matrix}\right.$

<=> $\left[\begin{matrix} x=\dfrac{\pi}{2}+k2\pi\\ x=x-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{3}-x+k2\pi\end{matrix}\right.$

<=> $\left[\begin{matrix} x=\dfrac{\pi}{2}+k2\pi\\ 0x=-\dfrac{\pi}{3}+k2\pi (VN)\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.$    (k in ZZ)

Vậy S={\pi/2+k2\pi;\pi/6+k\pi|k in ZZ}