\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+\left(sinx.cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2\left(cosx-sinx\right)+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\2\left(cosx-sinx\right)+sinx.cosx-1=0\end{matrix}\right.\)
TH1: \(sinx+cosx=0\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
TH2: \(2\left(cosx-sinx\right)+sinx.cosx-1=0\)
Đặt \(cosx-sinx=-\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=a\) (\(\left|a\right|\le\sqrt{2}\))
\(\Rightarrow a^2=1-2sinx.cosx\Rightarrow sinx.cosx=\frac{1-a^2}{2}\)
\(2a+\frac{1-a^2}{2}-1=0\)
\(\Leftrightarrow a^2-4a+1=0\Rightarrow\left[{}\begin{matrix}a=2+\sqrt{3}\left(l\right)\\a=2-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=2-\sqrt{3}\)
\(\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{3}-2}{\sqrt{2}}=sin\alpha\)
\(\Rightarrow...\)
Nghiệm thứ 2 xấu vậy, bạn có ghi đề nhầm chỗ nào ko nhỉ?
