Đáp án:
d) sin$(\dfrac{\pi}{6}-x)-sin^2$ $(\dfrac{\pi}{6}-x)=0(*)$
Đặt $sin(\dfrac{\pi}{6}-x)=t; t∈[-1;1]$
Từ (*), có: $t-t^2=0⇔$ \(\left[ \begin{array}{l}t=1 \\t=0\end{array} \right.\) (nhận)
Vì: $sin(\dfrac{\pi}{6}-x)=t$ $⇒$ \(\left[ \begin{array}{l}sin(\dfrac{\pi}{6}-x)=1(1)\\sin(\dfrac{\pi}{6}-x)=0(2)\end{array} \right.\)
Từ (1)⇒ \(\left[ \begin{array}{l}\pi/6-x=\pi/2+k2\pi\\\pi/6-x=\pi/2+k2\pi\end{array} \right.\) $⇔ x = -\pi/3+k2\pi$
Từ (2) ⇒ \(\left[ \begin{array}{l}\pi/6-x=k2\pi\\\pi/6-x=\pi+k2\pi\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\pi/6-k2\pi\\x=-5\pi/6+k2\pi\end{array} \right.\)
Vậy $S=${$-\pi/3+k2\pi; \pi/6-k2\pi; -5\pi/6+k2\pi;k∈Z$ }
BẠN THAM KHẢO NHA!!!
