Đáp án:
${\left[\begin{aligned}x=\dfrac{\arccos\dfrac{\sqrt{3}}{3}}{5}-\dfrac{1}{5}+\dfrac{k2\pi}{5}\\x=\dfrac{-\arccos\dfrac{\sqrt{3}}{3}}{5}-\dfrac{1}{5}+\dfrac{k2\pi}{5}\end{aligned}\right.},(k\in \mathbb{Z})\\$
Giải thích các bước giải:
$\cos(5x+1)=\dfrac{\sqrt{3}}{3}\\
\Leftrightarrow {\left[\begin{aligned}5x+1=\arccos\dfrac{\sqrt{3}}{3}+k2\pi\\5x+1=-\arccos\dfrac{\sqrt{3}}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}5x=\arccos\dfrac{\sqrt{3}}{3}-1+k2\pi\\5x=-\arccos\dfrac{\sqrt{3}}{3}-1+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{\arccos\dfrac{\sqrt{3}}{3}}{5}-\dfrac{1}{5}+\dfrac{k2\pi}{5}\\x=\dfrac{-\arccos\dfrac{\sqrt{3}}{3}}{5}-\dfrac{1}{5}+\dfrac{k2\pi}{5}\end{aligned}\right.},(k\in \mathbb{Z})\\$
