Bài 1:
a: \(cos\left(x+\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\)
=>\(\left[\begin{array}{l}x+\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\ x+\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k2\pi\\ x=-\frac23\pi+k2\pi\end{array}\right.\)
b:
ĐKXĐ: \(\begin{cases}x+30^0<>k\cdot180^0\\ 40^0-3x<>90^0+k\cdot180^0\end{cases}\Rightarrow\begin{cases}x<>-30^0+k\cdot180^0\\ 3x<>-50^0-k\cdot180^0\end{cases}\)
=>\(\begin{cases}x<>-30^0+k\cdot180^0\\ x<>-\frac{50^0}{3}-k\cdot60^0\end{cases}\)
\(\cot\left(x+30^0\right)=\tan\left(40^0-3x\right)\)
=>\(\tan\left(40^0-3x\right)=\tan\left(90^0-x-30^0\right)=\tan\left(60^0-x\right)\)
=>\(40^0-3x=60^0-x\)
=>\(3x-40^0=x-60^0\)
=>\(2x=-60^0+40^0=-20^0\)
=>\(x=-10^0\) (nhận)
c: \(2\cdot cos^2x-\sqrt2\cdot cosx-2=0\)
=>\(2\cdot cos^2x-2\sqrt2\cdot cosx+\sqrt2\cdot cosx-2=0\)
=>\(2\cdot cosx\left(cosx-\sqrt2\right)+\sqrt2\left(cosx-\sqrt2\right)=0\)
=>\(\left(cosx-\sqrt2\right)\left(2\cdot cosx+\sqrt2\right)=0\)
=>\(2\cdot cosx+\sqrt2=0\)
=>\(cosx=-\frac{1}{\sqrt2}\)
=>\(\left[\begin{array}{l}x=\frac34\pi+k2\pi\\ x=-\frac34\pi+k2\pi\end{array}\right.\)
d: \(cos\left(\frac{x}{4}\right)-\sqrt3\cdot\sin\left(\frac{x}{4}\right)=-1\)
=>\(\frac12\cdot cos\left(\frac{x}{4}\right)-\frac{\sqrt3}{2}\cdot\sin\left(\frac{x}{4}\right)=-\frac12\)
=>\(cos\left(\frac{x}{4}-\frac{\pi}{6}\right)=-\frac12\)
=>\(\left[\begin{array}{l}\frac{x}{4}-\frac{\pi}{6}=\frac23\pi+k2\pi\\ \frac{x}{4}-\frac{\pi}{6}=-\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{4}=\frac{\pi}{6}+\frac23\pi+k2\pi=\frac56\pi+k2\pi\\ \frac{x}{4}=-\frac23\pi+\frac{\pi}{6}+k2\pi=-\frac46\pi+\frac{\pi}{6}+k2\pi=-\frac12\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{10}{3}\pi+k8\pi\\ x=-2\pi+k8\pi\end{array}\right.\)
