Ngân hàng bài tập
A
Phương trình $\sin x+\sqrt{3}\cos x=\sqrt{2}$ có nghiệm $x=\alpha+k2\pi$ và $x=\beta+k2\pi$ với $-\dfrac{\pi}{2}<\alpha,\,\beta<\dfrac{\pi}{2}$ $(k\in\mathbb{Z})$. Khi đó, $\alpha\cdot\beta$ bằng
 | $\dfrac{7\pi^2}{144}$ |
 | $-\dfrac{5\pi^2}{144}$ |
 | $\dfrac{5\pi^2}{144}$ |
 | $-\dfrac{7\pi^2}{144}$ |
1 lời giải
Chọn phương án B.
$\begin{aligned}
\sin x+\sqrt{3}\cos x=\sqrt{2}\Leftrightarrow&\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x=\dfrac{\sqrt{2}}{2}\\
\Leftrightarrow&\sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\\
\Leftrightarrow&\left[\begin{array}{l}
x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\
x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=-\dfrac{\pi}{12}+k2\pi \\
x=\dfrac{5\pi}{12}+k2\pi
\end{array}\right.\;(k\in\mathbb{Z}).
\end{aligned}$
Vì $-\dfrac{\pi}{2}<\alpha,\,\beta<\dfrac{\pi}{2}$ nên $\begin{cases}
\alpha=-\dfrac{\pi}{12}\\
\beta=\dfrac{5\pi}{12}.
\end{cases}$
\alpha=-\dfrac{\pi}{12}\\
\beta=\dfrac{5\pi}{12}.
\end{cases}$
Khi đó, $\alpha\cdot\beta=-\dfrac{\pi}{12}\cdot\dfrac{5\pi}{12}=-\dfrac{5\pi^2}{144}$.