i. sin3x = $\sqrt{3}$ (cos3x-1) j. sin5x = $\frac{1}{3}$ (3- $\sqrt{3}$cos5x)

Đáp án:

 $i)  {\left[\begin{aligned}x =\dfrac{k2\pi}{3}\\ x=\dfrac{5\pi}{9} +\dfrac{k2\pi}{3}\end{aligned}\right.}\\
j)
{\left[\begin{aligned}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\ x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{aligned}\right.},(k\in \mathbb{Z})\\$

Giải thích các bước giải:

 $i) \sin3x=\sqrt{3}(\cos3x-1)\\
\Leftrightarrow \sin3x-\sqrt{3}\cos3x=-\sqrt{3}\\
\Leftrightarrow \dfrac{1}{2}\sin3x-\dfrac{\sqrt{3}}{2}\cos3x=\dfrac{-\sqrt{3}}{2}\\
\Leftrightarrow \sin3x\cos\dfrac{\pi}{3}-\cos3x\sin\dfrac{\pi}{3}=\dfrac{-\sqrt{3}}{2}\\
\Leftrightarrow \sin\left ( 3x-\dfrac{\pi}{3} \right )=\dfrac{-\sqrt{3}}{2}\\
\Leftrightarrow  {\left[\begin{aligned}3x-\dfrac{\pi}{3} =\dfrac{-\pi}{3}+k2\pi\\ 3x-\dfrac{\pi}{3} =\pi+\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned}3x =\dfrac{-\pi}{3}+\dfrac{\pi}{3}+k2\pi\\ 3x=\pi+\dfrac{\pi}{3}+\dfrac{\pi}{3} +k2\pi\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned}3x =k2\pi\\ 3x=\dfrac{5\pi}{3} +k2\pi\end{aligned}\right.}\\
\Leftrightarrow  {\left[\begin{aligned}x =\dfrac{k2\pi}{3}\\ x=\dfrac{5\pi}{9} +\dfrac{k2\pi}{3}\end{aligned}\right.}\\
j)
\sin5x=\dfrac{1}{3}(3-\sqrt{3}\cos5x)\\
\Leftrightarrow \sin5x+\dfrac{1}{\sqrt{3}}\cos5x=1\\
\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin5x+\dfrac{1}{2}\cos5x=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow \sin5x\cos\dfrac{\pi}{6}+\cos5x\sin\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow \sin\left (  5x+\dfrac{\pi}{6}\right )=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow {\left[\begin{aligned}5x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\ 5x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\ 5x=\pi-\dfrac{\pi}{6}-\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}5x=\dfrac{\pi}{6}+k2\pi\\ 5x=\dfrac{\pi}{2}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\ x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{aligned}\right.},(k\in \mathbb{Z})\\$