1.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pm\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow cos^2x-6sinx.cosx+sin^2x=-2\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(1-6tanx+tan^2x=-\frac{2}{cos^2x}\)
\(\Leftrightarrow tan^2x-6tanx+1=-2\left(1+tan^2x\right)\)
\(\Leftrightarrow3tan^2x-6tanx+3=0\)
\(\Leftrightarrow3\left(tanx-1\right)^2=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)