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36.
\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)
\(\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
37.
\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)
38.
\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)
39.
\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
33.
\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)
34.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)
35.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)
\(\Leftrightarrow x\ne k\pi\)
Với hàm \(y=f\left(x\right)=cot4x\)
Ta có \(f\left(-x\right)=cot\left(-4x\right)=-cot4x=-f\left(x\right)\)
\(\Rightarrow y=cot4x\) là hàm lẻ
d/
\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)
c/
\(\Leftrightarrow sinx-sin^2x=cosx-cos^2x\)
\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow sinx-cosx-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\1-sinx-cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\1-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
Với hàm \(y=f\left(x\right)=cosx\)
Ta thấy \(f\left(-x\right)=cos\left(-x\right)=cosx=f\left(x\right)\)
Nên \(y=cosx\) là hàm chẵn
Xét hàm \(y=f\left(x\right)=sin2x\)
\(f\left(-x\right)=sin\left(-2x\right)=-sin2x=-f\left(x\right)\)
\(\Rightarrow y=sin2x\) là hàm lẻ