równania trygonometryczne

[BarT123oks]

1) \(2\cos x\tg x-{2\sqrt3\over3}\cos x+\sqrt3\tg x=1,\ x\in [-\pi;\pi]\)
2) \(\cos x+|\cos x|=\sin2x,\ x \in [0,2\pi]\)

[Jerry]

2) \(\cos x+|\cos x|=\sin2x,\ x \in [0,2\pi]\)

\[\begin{cases}\cos x<0\\ \sin2x=0\end{cases}\vee\begin{cases}\cos x\ge0\\ 2\cos x=2\sin x\cos x\end{cases}\\
\begin{cases}\cos x<0\\ \sin x=0\vee \cos x=0\end{cases}\vee\begin{cases}\cos x\ge0\\ \cos x=0\vee \sin x=1\end{cases}\\
\cos x=-1\vee \cos x =0\\
x=\pi\vee (x={\pi\over2}\vee x={3\pi\over2})\]
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PS. graficznie

[Jerry]

1) \(2\cos x\tg x-{2\sqrt3\over3}\cos x+\sqrt3\tg x=1,\ x\in [-\pi;\pi]\)

Dla \(\cos x\ne 0\) mamy
\[2\cos x\left(\tg x -{\sqrt3\over3}\right)+\sqrt3\left(\tg x -{\sqrt3\over3}\right)=0\\
\tg x ={\sqrt3\over3}\vee \cos x=-{\sqrt3\over2}\\
(x=-{5\pi\over6}\vee x={\pi\over6})\vee (x=-{5\pi\over6}\vee x={5\pi\over6})\]
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