Question

Giải các phương trình :

a, \(cos\left(2x+\frac{\pi}{6}\right)=0\)

b, \(cos\left(4x-\frac{\pi}{3}\right)=1\)

c, \(cos\left(2x+25^o\right)=\frac{-\sqrt{2}}{2}\)

d, cot\(\left(3x+10^o\right)=\frac{\sqrt{3}}{3}\)

Answer 1

a/ \(cos\left(2x+\frac{\pi}{6}\right)=0\)

\(\Leftrightarrow2x+\frac{\pi}{6}=\frac{\pi}{2}+k\pi\)

\(\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

b/ \(cos\left(4x-\frac{\pi}{3}\right)=1\)

\(\Leftrightarrow4x-\frac{\pi}{3}=k2\pi\)

\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{2}\)

c/ \(cos\left(2x+25^0\right)=-\frac{\sqrt{2}}{2}=cos135^0\)

\(\Rightarrow\left[{}\begin{matrix}2x+25^0=135^0+k360^0\\2x+25^0=-135^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=55^0+k180^0\\x=-80^0+k180^0\end{matrix}\right.\)

d/ \(cot\left(3x+10^0\right)=\frac{\sqrt{3}}{3}=cot60^0\)

\(\Rightarrow3x+10^0=60^0+k180^0\)

\(\Rightarrow x=\frac{50^0}{3}+k60^0\)