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e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=2x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\pi-2x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
ĐKXĐ: ....
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6sin^2\left(x-\frac{\pi}{8}\right)}{cos^2\left(x-\frac{\pi}{8}\right)}\)
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6\left(1-cos\left(2x-\frac{\pi}{4}\right)\right)}{1+cos\left(2x-\frac{\pi}{4}\right)}\)
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6\left(\sqrt{2}-\left(sin2x+cos2x\right)\right)}{\sqrt{2}+sin2x+cos2x}\)
Đặt \(sin2x+cos2x=a\Rightarrow sin4x=a^2-1\)
\(\frac{1-2\sqrt{2}a}{a^2-1}=\frac{6\sqrt{2}-6a}{\sqrt{2}+a}\Leftrightarrow6a^3-8\sqrt{2}a^2-9a+7\sqrt{2}=0\)
\(\Leftrightarrow\left(2a-\sqrt{2}\right)\left(6a^2-5\sqrt{2}a-14\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{\sqrt{2}}{2}\\6a^2-5\sqrt{2}a-14=0\end{matrix}\right.\)
Nghiệm sau dị thật
Từ phương trình ban đầu ta có : \(2\cos5x\sin x=\sqrt{3}\sin^2x+\sin x\cos x\)
\(\Leftrightarrow\begin{cases}\sin x=0\\2\cos5x=\sqrt{3}\sin x+\cos x\end{cases}\)
+) \(\sin x=0\Leftrightarrow x=k\pi\)
+)\(2\cos5x=\sqrt{3}\sin x+\cos x\Leftrightarrow\cos5x=\cos\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{18}+\frac{k\pi}{3}\end{cases}\)