Bài 1:
1: \(\sin\left(10x+10^0\right)=\sin7x\)
=>\(\left[\begin{array}{l}10x+10^0=7x+k\cdot360^0\\ 10x+10^0=180^0-7x+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-10^0+k\cdot360^0\\ 17x=170^0+k\cdot360^0\end{array}\right.\)
=>\(\left[\begin{array}{l}x=-\frac{10}{3}^0+k\cdot120^0\\ x=10^0+k\cdot\frac{360^0}{17}\end{array}\right.\)
2: \(cos8x=-\frac{\sqrt3}{2}\)
=>\(\left[\begin{array}{l}8x=\frac56\pi+k2\pi\\ 8x=-\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{5}{48}\pi+\frac{k\pi}{4}\\ x=-\frac{5}{48}\pi+\frac{k\pi}{4}\end{array}\right.\)
3:
ĐKXĐ: \(2x+\frac{\pi}{5}<>\frac{\pi}{2}+k\pi\)
=>\(2x<>\frac{3}{10}\pi+k\pi\)
=>\(x<>\frac{3}{20}\pi+\frac{k\pi}{2}\)
\(\tan\left(2x+\frac{\pi}{5}\right)=5\)
=>\(2x+\frac{\pi}{5}=\arctan\left(5\right)+k\pi\)
=>\(2x=-\frac{\pi}{5}+\arctan\left(5\right)+k\pi\)
=>\(x=-\frac{\pi}{10}+\frac12\cdot\arctan\left(5\right)+\frac{k\pi}{2}\)
4:
ĐKXĐ: \(\begin{cases}4x<>k\pi\\ 2x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow\begin{cases}x<>\frac{k\pi}{4}\\ x<>\frac{\pi}{4}+k\pi\end{cases}\)
=>\(x<>\frac{k\pi}{4}\)
cot 4x=tan 2x
=>\(\tan2x=\tan\left(\frac{\pi}{2}-4x\right)\)
=>\(2x=\frac{\pi}{2}-4x+k\pi\)
=>\(6x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{12}+\frac{k\pi}{6}\)
