Question

tìm tập xác định

a) y=\(\sqrt{1-cos^2x}\)

b) y= \(\dfrac{sinx}{cos\left(x-\pi\right)}\)

c) y=\(\dfrac{1}{tanx-1}\)

Answer 1

a)
\(1-cos^2x\ge0\)

\(cos^2x\le1\)

\(\frac{1+cos2x}{2}\)\(\le1\)

1 + cos2x \(\le\)2

cos2x \(\le1\)

2x \(\le\)k2π

x ≤ kπ

b)

cos (x-π) ≠ 0

x-π ≠ \(\frac{\pi}{2}\) +kπ

x ≠ \(\frac{3\pi}{2}\) +kπ

c) \(\left\{{}\begin{matrix}cosx\ne0\\tanx-1\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\tanx\ne tan\frac{\pi}{4}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\text{​​}\ne\frac{\pi}{2}+k\pi\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)