Đáp án:
A: $\left\{\begin{array}{I}x=\dfrac{\pi}8+\dfrac{k\pi}2\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}2\end{array}\right.$ $(k\in\mathbb Z)$
B: $x =\dfrac{ \pi}6$
Lời giải:
A:
$2\sin\left({4x-\dfrac{\pi}3}\right)-1=0$
$\Leftrightarrow \sin\left({4x-\dfrac{\pi}3}\right)=\dfrac12$
$\Leftrightarrow\left[\begin{array}{I}4x-\dfrac{\pi}3=\dfrac{\pi}6+k2\pi\\4x-\dfrac{\pi}3=\pi-\dfrac{\pi}6+k2\pi\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{I}x=\dfrac{\pi}8+\dfrac{k\pi}2\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}2\end{array}\right.$ $(k\in\mathbb Z)$
B:
$2\sin^2x-3\sin x+1=0$
$\Leftrightarrow \sin x = 1$ hoặc $\sin x =\dfrac 12$
$\Leftrightarrow x =\dfrac{ \pi}2 + 2k\pi$ hoặc $x =\dfrac{ \pi}6 + 2k\pi$ hoặc $x =\dfrac{ 5\pi}6 + 2k\pi$. $(k\in\mathbb Z)$
Do $0 \leq x < \dfrac{\pi}2$ nên $x = \dfrac{\pi}6$.
