Mam problem z tymi rownaniami:
a) \(\displaystyle{ 4(log_2cosx)^2 + log_2(1+cos2x)=3}\)
b) \(\displaystyle{ sin(\pi\ logx) + cos(\pi\ logx)=1}\)
\(\displaystyle{ \frac{1}{2} ^{log _{0.5} ^{2}sinx } + (sinx) ^{log _{0.5} sinx} =1}\)
Prosilbym o pomoc w rozwiazaniu!
z gory dziekuje
Rozwiązywanie równań trygonometrycznych
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- Użytkownik
- Posty: 1
- Rejestracja: 23 lis 2008, o 17:43
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- tomekture8
- Użytkownik
- Posty: 194
- Rejestracja: 13 sty 2008, o 21:38
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- Podziękował: 21 razy
- Pomógł: 40 razy
Rozwiązywanie równań trygonometrycznych
w podpunkcjie b podstawiasz za
\(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = t
\(\displaystyle{ \sin t + \cos t = 1}\)
Po obliczeniu wychodzi
t = \(\displaystyle{ 2k \pi}\) lub t = \(\displaystyle{ \frac{\pi}{2} + 2k \pi}\)
\(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = \(\displaystyle{ 2k \pi}\) lub \(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = \(\displaystyle{ \frac{\pi}{2} + 2k \pi}\)
podpunkt a
4\(\displaystyle{ (\log_{2}\cos x) ^{2}}\) = \(\displaystyle{ (2\log_{2}\cos x) ^{2}}\) = \(\displaystyle{ [\log_{2}(\cos x) ^{2} ] ^{2}}\)
\(\displaystyle{ 1+ \cos 2x = 1 + (\cos x) ^{2} - (\sin x) ^{2} = 1 + (\cos x) ^{2} - 1 + (\cos x) ^{2} = 2(\cos x) ^{2}}\)
\(\displaystyle{ \log_{2} (1+ \cos 2x) = \log_{2} [2(\cos x) ^{2}] = \log_{2} [(\cos x) ^{2}] + \log_{2} 2}\)
Ostatecznie:
\(\displaystyle{ [\log_{2}(\cos x) ^{2} ] ^{2}}\) + \(\displaystyle{ \log_{2} [(\cos x) ^{2}] + \log_{2} 2 - 3 = 0}\)
\(\displaystyle{ \log_{2} [(\cos x) ^{2}] = t}\)
\(\displaystyle{ t^{2} + t - 2 = 0}\)
podpunkt c
\(\displaystyle{ 0.5^{\log^{2}_{0.5} \sin x} = (0.5^{ \log_{0.5} sin x }) ^{\log_{0.5} sin x} = (\sin x) ^{\log_{0.5} sin x}}\)
\(\displaystyle{ (\sin x) ^{\log_{0.5} sin x} + (\sin x) ^{\log_{0.5} sin x} = 1}\)
\(\displaystyle{ (\sin x) ^{\log_{0.5} sin x} = \frac{1}{2}}\)
\(\displaystyle{ \log_{0.5}(\sin x ) ^{{\log_{0.5} sin x}} = \log_{0.5} \frac{1}{2}}\)
\(\displaystyle{ {\log_{0.5} sin x} * {\log_{0.5} sin x} = 1}\)
\(\displaystyle{ {\log_{0.5} sin x} = 1}\)
\(\displaystyle{ \sin x = \frac{1}{2}}\)
\(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = t
\(\displaystyle{ \sin t + \cos t = 1}\)
Po obliczeniu wychodzi
t = \(\displaystyle{ 2k \pi}\) lub t = \(\displaystyle{ \frac{\pi}{2} + 2k \pi}\)
\(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = \(\displaystyle{ 2k \pi}\) lub \(\displaystyle{ \pi}\)\(\displaystyle{ \log (x)}\) = \(\displaystyle{ \frac{\pi}{2} + 2k \pi}\)
podpunkt a
4\(\displaystyle{ (\log_{2}\cos x) ^{2}}\) = \(\displaystyle{ (2\log_{2}\cos x) ^{2}}\) = \(\displaystyle{ [\log_{2}(\cos x) ^{2} ] ^{2}}\)
\(\displaystyle{ 1+ \cos 2x = 1 + (\cos x) ^{2} - (\sin x) ^{2} = 1 + (\cos x) ^{2} - 1 + (\cos x) ^{2} = 2(\cos x) ^{2}}\)
\(\displaystyle{ \log_{2} (1+ \cos 2x) = \log_{2} [2(\cos x) ^{2}] = \log_{2} [(\cos x) ^{2}] + \log_{2} 2}\)
Ostatecznie:
\(\displaystyle{ [\log_{2}(\cos x) ^{2} ] ^{2}}\) + \(\displaystyle{ \log_{2} [(\cos x) ^{2}] + \log_{2} 2 - 3 = 0}\)
\(\displaystyle{ \log_{2} [(\cos x) ^{2}] = t}\)
\(\displaystyle{ t^{2} + t - 2 = 0}\)
podpunkt c
\(\displaystyle{ 0.5^{\log^{2}_{0.5} \sin x} = (0.5^{ \log_{0.5} sin x }) ^{\log_{0.5} sin x} = (\sin x) ^{\log_{0.5} sin x}}\)
\(\displaystyle{ (\sin x) ^{\log_{0.5} sin x} + (\sin x) ^{\log_{0.5} sin x} = 1}\)
\(\displaystyle{ (\sin x) ^{\log_{0.5} sin x} = \frac{1}{2}}\)
\(\displaystyle{ \log_{0.5}(\sin x ) ^{{\log_{0.5} sin x}} = \log_{0.5} \frac{1}{2}}\)
\(\displaystyle{ {\log_{0.5} sin x} * {\log_{0.5} sin x} = 1}\)
\(\displaystyle{ {\log_{0.5} sin x} = 1}\)
\(\displaystyle{ \sin x = \frac{1}{2}}\)