Đáp án:
Giải thích các bước giải:
$a)sin(2x-\dfrac{\pi}{3})=\dfrac{1}{2}\\\Leftrightarrow sin(2x-\dfrac{\pi}{3})= sin(\dfrac{\pi}{6})\\\Leftrightarrow\left[\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\\\Leftrightarrow\left[\begin{matrix}x=\dfrac{\pi}{4}+ k\pi\\ x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.(k\in Z)$
$b)tan4x=\dfrac{1}{2}\\\Leftrightarrow4x=arctan\dfrac{1}{2}+k\pi\\\Leftrightarrow x=\dfrac{1}{4}arctan\dfrac{1}{2}+k\dfrac{\pi}{4} (k \in Z) $
$c) tan5x + cot2x=0\\ĐKXĐ:\left \{ {{cos5x\not= 0 } \atop {sin2x\not = 0}}\right.\Leftrightarrow \left \{{x\not= \dfrac{\pi}{10}+ k\dfrac{\pi}{5}\atop{x \not = k\dfrac{\pi}{2}}} \right. (k \in Z) \\\Leftrightarrow tan5x=-cot2x\\\Leftrightarrow tan5x = tan (\dfrac{\pi}{2}+2x)\\\Leftrightarrow 5x = \dfrac{\pi}{2}+2x+k\pi\\\Leftrightarrow x = \dfrac{\pi}{6} + k \dfrac{\pi}{3} (k \in Z)$
