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\(\frac{1+sin2a}{1-sin2a}=\frac{sin^2a+cos^2a+2sina.cosa}{sin^2a+cos^2a-2sina.cosa}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)^2}\)
\(=\frac{\left(\sqrt{2}cos\left(a-\frac{\pi}{4}\right)\right)^2}{\left(\sqrt{2}sin\left(a-\frac{\pi}{4}\right)\right)^2}=\frac{cos^2\left(a-\frac{\pi}{4}\right)}{sin^2\left(a-\frac{\pi}{4}\right)}=cot^2\left(a-\frac{\pi}{4}\right)\)
d/
\(\Leftrightarrow sin2x=sin6x-sin4x\)
\(\Leftrightarrow2sinx.cosx=2cos5x.sinx\)
\(\Leftrightarrow sinx\left(cosx-cos5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos5x=cosx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\5x=x+k2\pi\\5x=-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\)
a/ Bạn coi lại vế trái đề bài, nhìn không hợp lý
b/ \(\Leftrightarrow\frac{1}{2}sin9x-\frac{1}{2}sinx=\frac{1}{2}sin5x-\frac{1}{2}sinx\)
\(\Leftrightarrow sin9x=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}9x=5x+k2\pi\\9x=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k\pi}{7}\end{matrix}\right.\)
c/ \(\Leftrightarrow sin2x-cos2x=cosx-sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow cos\left(\frac{3\pi}{4}-2x\right)=cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{3\pi}{4}-2x=x+\frac{\pi}{4}+k2\pi\\\frac{3\pi}{4}-2x=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)
