Rownanie trygonometryczne

[lukd]

\(\displaystyle{ 4( log _{2} cosx) ^{2} +log _{2} (1+cos2x)=3}\)

[lorakesz]

\(\displaystyle{ 4( log _{2} cosx) ^{2} +log _{2} (1+cos2x)=3}\)

\(\displaystyle{ 4( \log _{2}\cos x) ^{2} +\log _{2} (1+\cos2x)=3\\
4( \log _{2}\cos x) ^{2} +\log _{2} (\sin^2x+\cos^2x+\cos^2x-\sin^2x)=3\\
4( \log _{2}\cos x) ^{2} +\log _{2} (2\cos^2x)=3\\
4( \log _{2}\cos x) ^{2} +2\log _{2}\cos x+1=3\\
2( \log _{2}\cos x) ^{2} +\log _{2}\cos x-1=0\\
t=\log_2\cos x, \quad t\in (-\infty, 0> \\
2t^{2} +t-1=0\\
(2t-1)(t+1)=0 t=-1\\
-1=\log_2\cos x
2^{-1}=\cos x\\
\frac{1}{2}=\cos x\\
x=\frac{\pi}{3}+2k\pi, k\in Z}\)

[lukd]

Mógłbyś mi napisać skąd wzięło się:
\(\displaystyle{ 2log _{2} cosx+1}\)

[lorakesz]

\(\displaystyle{ \log _{2} (2\cos^2x)=\log_22+\log_2\cos^2x=1+\log_2(\cos x )^2}\)
Ponieważ
\(\displaystyle{ \log_ab^c=c\log_ab}\)
to:
\(\displaystyle{ 1+\log_2(\cos x )^2=1+2\log_2(\cos x )}\)