Câu 16: TXĐ: $D=R$
$2sin^23x=1$
$↔ sin^23x=\dfrac{1}{2}$
$↔ sin3x=±\dfrac{\sqrt[]{2}}{2}$
Với $sin3x=\dfrac{\sqrt[]{2}}{2}$, ta có:
$\left[ \begin{array}{l}3x=\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{3\pi}{4}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\end{array} \right.$
Với $sin3x=-\dfrac{\sqrt[]{2}}{2}$, ta có:
$\left[ \begin{array}{l}3x=-\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{5\pi}{4}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{array} \right.$
Kết hợp nghiệm ta được nghiệm chung là:
$x=\dfrac{\pi}{12}+k\dfrac{\pi}{6}$ $(k∈Z)$
(Đáp án $D$)
