\(\left|cosx\right|-\left|sinx\right|-\left(\left|cosx\right|-\left|sinx\right|\right)\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=0\)
\(\Leftrightarrow\left(\left|cosx\right|-\left|sinx\right|\right)\left(1-\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|cosx\right|=\left|sinx\right|\Leftrightarrow cos2x=0\left(1\right)\\\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
\(\left(2\right)\Leftrightarrow\left|cosx\right|+\left|sinx\right|=\dfrac{1}{\sqrt{1+sin2x}}\) (với \(sin2x\ne-1\))
\(\Leftrightarrow1+2\left|sinx.cosx\right|=\dfrac{1}{1+sin2x}\)
\(\Leftrightarrow1+\left|sin2x\right|=\dfrac{1}{1+sin2x}\)
TH1: \(-1< sin2x< 0\Rightarrow1-sin2x=\dfrac{1}{1+sin2x}\)
\(\Leftrightarrow1-sin^22x=1\Rightarrow sin2x=0\) (loại)
TH2: \(0\le sin2x\le1\Rightarrow1+sin2x=\dfrac{1}{1+sin2x}\)
\(\Leftrightarrow1+sin2x=1\Leftrightarrow sin2x=0\Rightarrow x=\dfrac{k\pi}{2}\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
Bạn tự tìm số giá trị nhé
