Question

Giải phương trình:

a) \(sin\left(\frac{\pi}{6}cosx+\frac{\pi}{3}\right)=0\)

b) \(cos\left(\pi cos3x\right)=0\)

c) \(tan\left(\frac{\pi}{3}sin\pi x\right)=\frac{1}{\sqrt{3}}\)

Answer 1

a/

\(\Leftrightarrow\frac{\pi}{6}cosx+\frac{\pi}{3}=k\pi\)

\(\Leftrightarrow cosx=-2+6k\)

Do \(-1\le cosx\le1\Rightarrow-1\le-2+6k\le1\)

\(\Rightarrow\frac{1}{6}\le k\le\frac{1}{2}\Rightarrow\) ko tồn tại k thỏa mãn

Vậy pt vô nghiệm

b.

\(\Leftrightarrow\pi cos3x=\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow cos3x=\frac{1}{2}+k\)

\(-1\le\frac{1}{2}+k\le1\Rightarrow k=\left\{-1;0\right\}\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 2

c. ĐKXĐ: ...

\(\Leftrightarrow\frac{\pi}{3}sin\pi x=\frac{\pi}{6}+k\pi\)

\(\Leftrightarrow sin\pi x=\frac{1}{2}+3k\)

\(-1\le\frac{1}{2}+3k\le1\Rightarrow k=0\)

\(\Rightarrow sin\pi x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\pi x=\frac{\pi}{6}+k2\pi\\\pi x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}+2k\\x=\frac{5}{6}+2k\end{matrix}\right.\)