Đáp án:

 $a)  x=\dfrac{11\pi}{30}+k\pi, k\in\mathbb{Z}\\
b)  {\left[\begin{aligned}x=\dfrac{1}{2}.\arcsin\dfrac{1}{3}-\dfrac{45^{\circ}}{2}+k\pi\\x=\dfrac{\pi-\arcsin\dfrac{1}{3}-45^{\circ}}{2}+k\pi\end{aligned}\right.},(k\in\mathbb{Z})\\
c) 
{\left[\begin{aligned}x=\dfrac{105^{\circ}}{2}+k180^{\circ}\\x=\dfrac{-15^{\circ}}{2}+k180^{\circ}\end{aligned}\right.},(k\in \mathbb{Z})\\
d)
{\left[\begin{aligned}x=\dfrac{7\pi}{36}-\dfrac{k2\pi}{3}\\x=\dfrac{-13\pi}{36}-\dfrac{k2\pi}{3}\end{aligned}\right.},(k\in \mathbb{Z})\\
e)
{\left[\begin{aligned}x=\dfrac{35^{\circ}}{2}+k\pi\\x=\dfrac{-65^{\circ}}{6}+\dfrac{k\pi}{3}\end{aligned}\right.},(K\in \mathbb{Z})\\$

Giải thích các bước giải:

 $a) 3\tan\left ( x-\dfrac{\pi}{5} \right )-\sqrt{3}=0\\
\Leftrightarrow 3\tan\left ( x-\dfrac{\pi}{5} \right )=\sqrt{3}\\
\Leftrightarrow \tan\left ( x-\dfrac{\pi}{5} \right )=\dfrac{\sqrt{3}}{3}\\
\Leftrightarrow x-\dfrac{\pi}{5} =\dfrac{\pi}{6}+k\pi\\
\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{\pi}{5}+k\pi\\
\Leftrightarrow x=\dfrac{11\pi}{30}+k\pi, k\in\mathbb{Z}\\
b) \sin(2x+45^{\circ})=\dfrac{1}{3}\\
\Leftrightarrow {\left[\begin{aligned}2x+45^{\circ}=\arcsin\dfrac{1}{3}+k2\pi\\2x+45^{\circ}=\pi-\arcsin\dfrac{1}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\arcsin\dfrac{1}{3}-45^{\circ}+k2\pi\\2x=\pi-\arcsin\dfrac{1}{3}-45^{\circ}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{2}.\arcsin\dfrac{1}{3}-\dfrac{45^{\circ}}{2}+k\pi\\x=\dfrac{\pi-\arcsin\dfrac{1}{3}-45^{\circ}}{2}+k\pi\end{aligned}\right.},(k\in\mathbb{Z})\\
c) 
2\cos(2x-45^{\circ})-1=0\\
\Leftrightarrow 2\cos(2x-45^{\circ})=1\\
\Leftrightarrow \cos(2x-45^{\circ})=\dfrac{1}{2}\\
\Leftrightarrow {\left[\begin{aligned}2x-45^{\circ}=60^{\circ}+k360^{\circ}\\2x-45^{\circ}=-60^{\circ}+k360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=60^{\circ}+45^{\circ}+k360^{\circ}\\2x=-60^{\circ}+45^{\circ}+k360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=105^{\circ}+k360^{\circ}\\2x=-15^{\circ}+k360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{105^{\circ}}{2}+k180^{\circ}\\x=\dfrac{-15^{\circ}}{2}+k180^{\circ}\end{aligned}\right.},(k\in \mathbb{Z})\\
d)
2\sin\left ( -3x+\dfrac{\pi}{4} \right )+\sqrt{3}=0\\
\Leftrightarrow 2\sin\left ( -3x+\dfrac{\pi}{4} \right )=-\sqrt{3}\\
\Leftrightarrow \sin\left ( -3x+\dfrac{\pi}{4} \right )=\dfrac{-\sqrt{3}}{2}\\
\Leftrightarrow {\left[\begin{aligned}-3x+\dfrac{\pi}{4} =\dfrac{-\pi}{3}+k2\pi\\-3x+\dfrac{\pi}{4} =\pi+\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}-3x=\dfrac{-\pi}{3}-\dfrac{\pi}{4} +k2\pi\\-3x=\pi-\dfrac{\pi}{4} +\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}-3x=\dfrac{-7\pi}{12}+k2\pi\\-3x=\dfrac{13\pi}{12}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7\pi}{36}-\dfrac{k2\pi}{3}\\x=\dfrac{-13\pi}{36}-\dfrac{k2\pi}{3}\end{aligned}\right.},(k\in \mathbb{Z})\\
e)
\cos(4x+50^{\circ})-\cos(2x+15^{\circ})=0\\
\Leftrightarrow \cos(4x+50^{\circ})=\cos(2x+15^{\circ})\\
\Leftrightarrow {\left[\begin{aligned}4x+50^{\circ}=2x+15^{\circ}+k2\pi\\4x+50^{\circ}=-2x-15^{\circ}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}4x-2x=15^{\circ}-50^{\circ}+k2\pi\\4x+2x=-15^{\circ}-50^{\circ}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=35^{\circ}+k2\pi\\6x=-65^{\circ}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{35^{\circ}}{2}+k\pi\\x=\dfrac{-65^{\circ}}{6}+\dfrac{k\pi}{3}\end{aligned}\right.},(K\in \mathbb{Z})\\$