Tìm các nghiệm

A

- Hướng dẫn giải

Ta có:

$$\begin{gathered} 4{\sin }^2\left(\mathrm{\pi }-\frac{\mathrm{x}}{2}\right)-\sqrt{3}\left(\frac{\mathrm{\pi }}{2}-2x\right) \\ =1+2{\cos }^2\left(x-\frac{3\mathrm{\pi }}{4}\right) \\ \iff 2\left[1-\cos \left(2\mathrm{\pi }-\mathrm{x}\right)\right] \\ -\sqrt{3}\cos \left(2x\right) \\ =1+1+\cos \left(2x-\frac{3\mathrm{\pi }}{2}\right) \\ \iff 2-2\cos \left(x\right)-\sqrt{3}\cos \left(2x\right) \\ =2-\sin \left(2x\right) \\ \iff \sin \left(2x\right)-\sqrt{3}\cos \left(2x\right) \\ =2\cos \left(x\right) \\ \iff \frac{1}{2}\sin \left(2x\right)-\frac{\sqrt{3}}{2}\cos \left(2x\right) \\ =\cos \left(x\right) \\ \iff \sin \left(2x-\frac{\mathrm{\pi }}{3}\right)=\cos \left(\frac{\mathrm{\pi }}{3}-x\right) \\ \iff \left\{\begin{array}{l}x=\frac{5\mathrm{\pi }}{8}+k\frac{2\mathrm{\pi }}{3}\\ x=\frac{5\mathrm{\pi }}{6}+k2\mathrm{\pi }\end{array}\right. \end{gathered}$$

Vì $x\in \left(0;\frac{\mathrm{\pi }}{2}\right)$ nên ta chọn được nghiệm $x=\frac{5\mathrm{\pi }}{8}$

Đáp án A