1.
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
2.
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
4.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{1}{4}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=\dfrac{1}{4}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{1}{4}\)
\(\Leftrightarrow cos4x=-1\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
5.
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=cos2x\)
\(\Leftrightarrow cos2x\left[\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x\right]=cos2x\)
\(\Leftrightarrow cos2x\left(1-\dfrac{1}{4}sin^22x\right)=cos2x\)
\(\Leftrightarrow cos2x\left(1-\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)\right)=cos2x\)
\(\Leftrightarrow cos2x\left(\dfrac{7}{8}+\dfrac{1}{8}cos4x\right)=cos2x\)
\(\Leftrightarrow cos2x\left(\dfrac{1}{8}cos4x-\dfrac{1}{8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\4x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
6.
\(\Leftrightarrow\pi.sinx=k2\pi\)
\(\Leftrightarrow sinx=2k\) (1)
Do \(-1\le sinx\le1\)
\(\Rightarrow-1\le2k\le1\)
\(\Rightarrow-\dfrac{1}{2}\le k\le\dfrac{1}{2}\)
\(\Rightarrow k=0\)
Thế vào (1):
\(\Rightarrow sinx=0\)
\(\Rightarrow x=n\pi\)
