Муодилаи тригонометриро ҳал кунед:
\(8\cos z\cos(60^{\circ}-z)\cos(60^{\circ}+z)+1=0\).
Ҳал.
\(8\cos z\cos(60^{\circ}-z)\cos(60^{\circ}+z)+1=0\)
\(4\cos z\cdot2\cos(60^{\circ}-z)\cos(60^{\circ}+z)+1=0\)
\(4\cos z\cdot2\cos\frac{120^{\circ}-2z}{2}\cos\frac{120^{\circ}+2z}{2}+1=0\)

\(\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\).
Яъне,
\(2\cos\frac{120^{\circ}-2z}{2}\cos\frac{120^{\circ}+2z}{2}=2\cos\frac{120^{\circ}+2z}{2}\cos\frac{120^{\circ}-2z}{2}=\)
\(=\cos120^{\circ}+\cos2z=\cos2z+\cos120^{\circ}\).

\(4\cos z\cdot(\cos2z+\cos120^{\circ})+1=0\)
\(4\cos z\cdot(\cos2z-\frac{1}{2})+1=0\)
\(4\cos z\cos2z-2\cos z+1=0\)
\(2\cdot2\cos z\cos2z-2\cos z+1=0\)

\(2\cos z\cos2z = 2\cos2z\cos z = 2\cos\frac{3z+z}{2}\cos\frac{3z-z}{2} =\)
\(=\cos3z+\cos z\).
\(2\cdot(\cos3z+\cos z)-2\cos z+1=0\)
\(2\cos3z+2\cos z-2\cos z+1=0\)
\(2\cos3z+1=0\)
\(2\cos3z=-1\)
\(\cos3z=-\frac{1}{2}\)
\(3z=\pm\arccos{-\frac{1}{2}}+2\pi n\),\(\quad n\in \mathbb{Z}\)
\(3z=\pm\frac{2\pi}{3}+2\pi n\),\(\quad n\in \mathbb{Z}\quad|\cdot\frac{1}{3}\)
\(z=\pm\frac{2\pi}{9}+\frac{2\pi n}{3}\),\(\quad n\in \mathbb{Z}\)
\(z=\pm40^{\circ}+\frac{2\pi}{3}n\),\(\quad n\in \mathbb{Z}\)
\(z=\pm40^{\circ}+120^{\circ}n\),\(\quad n\in \mathbb{Z}\).
Ҷавоб: \(z=\pm40^{\circ}+120^{\circ}n\),\(\quad n\in \mathbb{Z}\).