a.
\(\Leftrightarrow2cos^2\left(\pi cos^2x\right)-1=cos\left(\pi sin2x\right)\)
\(\Leftrightarrow cos\left(2\pi cos^2x\right)=cos\left(\pi sin2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2\pi cos^2x=\pi sin2x+k2\pi\\2\pi cos^2x=-\pi sin2x+n2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2cos^2x-sin2x=2k\\2cos^2x+sin2x=2n\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x-sin2x=2k-1\\cos2x+sin2x=2n-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{2k-1}{\sqrt{2}}\\cos\left(2x-\dfrac{\pi}{4}\right)=\dfrac{2n-1}{\sqrt{2}}\end{matrix}\right.\)
Do \(-1\le cosa\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{2k-1}{\sqrt{2}}\le1\\-1\le\dfrac{2n-1}{\sqrt{2}}\le1\end{matrix}\right.\) \(\Rightarrow k;n=\left\{0;1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}cos\left(2x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\cos\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\\cos\left(2x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow....\)
c.
Đặt \(2x+\dfrac{\pi}{8}=t\Rightarrow2x=t-\dfrac{\pi}{8}\Rightarrow4x=2t-\dfrac{\pi}{4}\)
Pt trở thành:
\(\sqrt{2}cos\left(2t-\dfrac{\pi}{4}\right)+4\sqrt{2}cost+1=0\)
\(\Leftrightarrow cos2t+sin2t+4\sqrt{2}cost+1=0\)
\(\Leftrightarrow2cos^2t+2sint.cost+4\sqrt{2}cost=0\)
\(\Leftrightarrow2cost\left(cost+sint+2\sqrt{2}\right)=0\)
\(\Leftrightarrow cost\left(\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)+2\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\cos\left(t+\dfrac{\pi}{4}\right)=-2\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{8}\right)=0\)
\(\Leftrightarrow...\)
