a/
\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)
\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)
\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)
b/
\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x=-\frac{1}{2}\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
c/
\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)
\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)
\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
d/
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)
\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)
\(\Leftrightarrow cos4x=1\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
Câu cuối bạn coi lại đề, sao 2 số hạng đầu giống hệt nhau vậy?
e/
\(2cos^2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow1+cos2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow2cos^32x+cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x-1\right)\left(2cos^22x+3cos2x+2\right)=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
