Đặt \(z=\cos x+i\sin x,x\in\left[0,2\pi\right]\)
\(1=\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right|=\frac{\left|z^2+\overline{z}^2\right|}{\left|z\right|^2}\)
\(=\left|\cos2x+i\sin2x+\cos2x-i\sin2x\right|\)
\(=2\left|\cos2x\right|\)
Do đó : \(\cos2x=\frac{1}{2}\) hoặc \(\cos2x=-\frac{1}{2}\)
- Nếu \(\cos2x=\frac{1}{2}\)
thì : \(x_1=\frac{\pi}{6},x_2=\frac{5\pi}{6},x_3=\frac{7\pi}{6},x_4=\frac{11\pi}{6}\)
- Nếu \(\cos2x=-\frac{1}{2}\)
thì : \(x_5=\frac{\pi}{3},x_6=\frac{2\pi}{3},x_7=\frac{4\pi}{3},x_8=\frac{5\pi}{3}\)
Do đó có 8 nghiệm :\(z_k=\cos x_k+i\sin x_kk=1,2,3,.....,8\)