równania trygonometryczne

[BarT123oks]

1) \((4-\cos(x))^2=8\sin^2(\frac{x}{2})-8\cos^2(\frac{x}{2})+\frac{65}{4}\), dla \(x \in (-\pi,\pi)\)
2) \((\sin(x)-1)^2=\frac{3}{2}-4\sin(\frac{x}{2})\cos\frac{x}{2}\), dla \(x \in (0, 2\pi)\)

[Jerry]

2) \((sin(x)-1)^2=\frac{3}{2}-4sin(\frac{x}{2})cos\frac{x}{2}\), dla \(x \in (0, 2\pi)\)

Równanie jest równoważne
\((\sin x-1)^2=\frac{3}{2}-2\sin x\\
\sin^2x={1\over2}\\
1-2\sin^2x=0\\
\cos2x=0\\
2x={\pi\over2}+k\cdot\pi\wedge k\in\zz\\
x={\pi\over4}+k\cdot{\pi\over2}\wedge k\in\zz\\
x \in (0, 2\pi)\So x\in\{{\pi\over4},{3\pi\over4},{5\pi\over4},{7\pi\over4}\}
\)

[Jerry]

1) \((4-cos(x))^2=8sin^2(\frac{x}{2})-8cos^2(\frac{x}{2})+\frac{65}{4}\), dla \(x \in (-\pi,\pi)\)

\((4-\cos x)^2=8\sin^2(\frac{x}{2})-8\cos^2(\frac{x}{2})+\frac{65}{4}\\
16-8\cos x+\cos^2x=-8\cos x+{65\over4}\\
\cos^2x={1\over4}\\
2\cos^2x-1=-{1\over2}\\
\cos2x=-{1\over2}\\
(2x={2\pi\over3}+k\cdot2\pi\vee2x=-{2\pi\over3}+k\cdot2\pi)\wedge k\in\zz\\
(x={\pi\over3}+k\cdot\pi\vee x=-{\pi\over3}+k\cdot\pi)\wedge k\in\zz\\
x \in (-\pi,\pi)\So x\in\{-{2\pi\over3},-{\pi\over3},{2\pi\over3},{\pi\over3}\}\)

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