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Ta có:
`@-1 <= sin x <= 1`
`<=>0 <= 1+sin x <= 2=>1+sin x >= 0`
`@-1 <= cos x <= 1`
`<=>1 >= -cos x >= -1`
`<=>2 >= 1-cos x >= 0=>1-cos x >= 0`
Hàm số xác định `<=>[1+sin x]/[1-cos x] >= 0`
`<=>{(1+sin x >= 0(L Đ)),(1-cos x > 0):}<=>1-cos x ne 0<=>x ne k2\pi (k in ZZ)`
`=>TXĐ: D=R\\{k2\pi| k in ZZ}`.
1.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx-sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\\dfrac{sinx}{cosx}-sinx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
2.
ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
3.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\cos\left(x-\dfrac{\pi}{4}\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{2}\right)\ne0\Leftrightarrow cos2x\ne0\)
\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
ĐK: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\sin^4x-cos^4x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{\pi}{4}\ne k\pi\\\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\cos2x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\2x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne1\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
a: ĐKXĐ: \(cosx-1\ne0\)
=>\(cosx\ne1\)
=>\(x\ne k2\Omega\)
b: ĐKXĐ: sin x-1>=0
=>sin x>=1
mà \(-1< =sinx< =1\)
nên sin x=1
=>\(x=\dfrac{\Omega}{2}+k2\Omega\)
c:
-1<=sin x<=1
=>-1+1<=sin x+1<=1+1
=>0<=sin x+1<=2
ĐKXĐ: \(\dfrac{1+sinx}{1-cosx}>=0\)
mà \(1+sinx>=0\)(cmt)
nên \(1-cosx>0\)
=>\(cosx< 1\)
mà -1<=cosx<=1
nên \(cosx\ne1\)
=>\(x\ne k2\Omega\)
\(y=\dfrac{cos3x}{1-sinx}+tanx=\dfrac{cos3x}{1-sinx}+\dfrac{sinx}{cosx}\)
Hàm số xác định khi \(\left\{{}\begin{matrix}1-sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinx\ne1\\cosx\ne0\end{matrix}\right.\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)