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Find the angle satisfying the following equation
\begin{aligned} &4 \log _{16}(\cos 2 x)+2 \log _{4}(\sin x) \\ &+\log _{2}(\cos x)+3=0 \end{aligned}
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## 1 Answer

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Best answer
$\cos 2 x=\cos ^{2} x-\sin ^{2} x$
$\cos x=a$
$\sin x=b$

$4 \log _{2}\left(a^{2}-b^{2}\right)+2 \log _{2}(b)+\log _{2}(a)+\log _{2} 8=0$

$\log _{2}\left(8 a^{3} b-8 a b^{3}\right)=0$

$8 a^{3} b-8 a b^{3}=2^{0}$

$8 a b\left(a^{2}-b^{2}\right)=1$

$8 \cos x \sin x \cos 2 x=1$

$4 \sin 2 x \cos 2 x=1$

$2 \sin 4 x=1$

$4 x=\sin { }^{-1}\left(\frac{1}{2}\right)$

$4 x=\frac{\pi}{6}$

$x=\frac{\pi}{24} \cdot \frac{180}{\pi}=\frac{15}{2}=7.5$

$4 x=\frac{5 \pi}{6}$

$x=\frac{5 \pi}{24} \cdot \frac{180}{\pi}=\frac{75}{2}=37.5$
by Diamond (71,587 points)

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