H24

giải phương trình

a) \(sinx=sin\dfrac{\pi}{4}\)

b) \(cos2x=cosx\)

c) \(tan\left(x-\dfrac{\pi}{3}\right)=\sqrt{3}\)

d) \(cot\left(2x+\dfrac{\pi}{6}\right)=cot\dfrac{\pi}{4}\)

NT
6 tháng 11 2023 lúc 19:20

a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)

b: cos2x=cosx

=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)

c:

ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)

=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)

 \(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)

=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{2}{3}\Omega+k\Omega\)

d:

ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)

=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)

=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)

 \(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)

=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)

=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)

=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)

Bình luận (0)