Giải phương trình $$\sin5x+\sqrt{3}\cos5x=2\sin\left(\dfrac{\pi}{2}-3x\right)$$
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\ x=\dfrac{\pi}{48}+k\dfrac{\pi}{4}\end{array}\right.\,(k \in \mathbb{Z})\) |
 | Vô nghiệm |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{12}+k\pi\\ x=\dfrac{\pi}{24}+k\dfrac{\pi}{4}\end{array}\right.\,(k \in \mathbb{Z})\) |
 | \(\left[\begin{array}{l}x=\dfrac{\pi}{12}+k\pi\\ x=\dfrac{\pi}{48}+k\dfrac{\pi}{4}\end{array}\right.\,(k \in \mathbb{Z})\) |
Chọn phương án D.
\(\begin{aligned}
&\sin5x+\sqrt{3}\cos5x=2\sin\left(\dfrac{\pi}{2}-3x\right)\\
\Leftrightarrow&\dfrac{1}{2}\sin5x+\dfrac{\sqrt{3}}{2}\cos5x=\sin\left(\dfrac{\pi}{2}-3x\right)\\
\Leftrightarrow&\sin\left(5x+\dfrac{\pi}{3}\right)=\sin\left(\dfrac{\pi}{2}-3x\right)\\
\Leftrightarrow&\left[\begin{array}{l}5x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-3x+k2\pi\\ 5x+\dfrac{\pi}{3}=\pi-\dfrac{\pi}{2}+3x+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}8x=\dfrac{\pi}{6}+k2\pi\\ 2x=\dfrac{\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{48}+k\dfrac{\pi}{4}\\ x=\dfrac{\pi}{12}+k\pi\end{array}\right.\,(k \in \mathbb{Z})
\end{aligned}\)