Question

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

Answer 1

\(\cos 2 x=\cos ^{2} x-\sin ^{2} x\)
\(\cos x=a\)
\(\sin x=b\)

 

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

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

\(8 a^{3} b-8 a b^{3}=2^{0}\)

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

\(8 \cos x \sin x \cos 2 x=1\)

\(4 \sin 2 x \cos 2 x=1\)

\(2 \sin 4 x=1\)

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

\(4 x=\frac{\pi}{6}\)

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

\(4 x=\frac{5 \pi}{6}\)

\(x=\frac{5 \pi}{24} \cdot \frac{180}{\pi}=\frac{75}{2}=37.5\)