Question

Giải phương trình:

a) \(tan\left(\frac{\pi}{2}sin\pi\left(x+1\right)\right)=1\)

b) \(tan\left(\frac{\pi}{3}cot\pi x\right)=\frac{1}{\sqrt{3}}\)

c) \(sin\left(\pi tan3x\right)=\frac{1}{2}\)

Answer 1

a/

\(\Leftrightarrow\frac{\pi}{2}sin\pi\left(x+1\right)=\frac{\pi}{4}+k\pi\)

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

Do \(-1\le sin\pi\left(x+1\right)\le1\Rightarrow k=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{6}+k2\pi\\\pi\left(x+1\right)=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

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

Answer 2

b.

ĐKXĐ: ...

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

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

\(\Leftrightarrow\pi x=arccot\left(\frac{1}{2}+3k\right)+n\pi\)

\(\Leftrightarrow x=\frac{1}{\pi}arccot\left(\frac{1}{2}+3k\right)+n\)

c.

\(\Leftrightarrow\left[{}\begin{matrix}\pi tan3x=\frac{\pi}{6}+k2\pi\\\pi tan3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tan3x=\frac{1}{6}+2k\\tan3x=\frac{5}{6}+2k\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}arctan\left(\frac{1}{6}+2k\right)+\frac{n2\pi}{3}\\x=\frac{1}{3}arctan\left(\frac{5}{6}+2k\right)+\frac{n2\pi}{3}\end{matrix}\right.\)