Lời giải:
Đề bài phải thêm đk về x. VD: \(x\in (-\frac{\pi}{2};0)\)
Ta có:
\(\sqrt{4\sin ^4x+\sin ^2(2x)}=\sqrt{4\sin ^4x+(2\sin x\cos x)^2}\)
\(=\sqrt{4\sin ^2x(\sin ^2x+\cos ^2x)}=\sqrt{4\sin ^2x}=|2\sin x|=-2\sin x\) do \(x\in (\frac{-\pi}{2};0)\)
Mặt khác:
\(\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)=\cos \frac{\pi}{4}\cos \frac{x}{2}+\sin \frac{\pi}{4}\sin \frac{x}{2}\)
\(=\frac{\sqrt{2}}{2}\cos \frac{x}{2}+\frac{\sqrt{2}}{2}\sin \frac{x}{2}\)
\(\Rightarrow 4\cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=2(\cos \frac{x}{2}+\sin \frac{x}{2})^2\)
\(=2(\cos ^2\frac{x}{2}+\sin ^2\frac{x}{2}+2\cos \frac{x}{2}\sin \frac{x}{2})\)
\(=2(1+\sin x)=2+2\sin x\)
Do đó: \(A=-2\sin x+2+2\sin x=2\) không phụ thuộc vào x
