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The general solution of $2 \cos 4 x+\sin ^2 2 x=0$ is
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Verified Answer
The correct answer is:
$x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right)$
Given, $2 \cos 4 x+\sin ^2 2 x=0$
$\begin{array}{ll}
\Rightarrow & 2 \cos 4 x+\left(\frac{1-\cos 4 x}{2}\right)=0 \\
\Rightarrow & 3 \cos 4 x+1=0
\end{array}$
$\begin{array}{ll}\Rightarrow & \cos 4 x=-\frac{1}{3} \\ \Rightarrow & \sin 4 x= \pm \frac{2 \sqrt{2}}{3} \\ \Rightarrow & 4 x=n \pi+(-1)^n \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right) \\ \Rightarrow & x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right)\end{array}$
$\begin{array}{ll}
\Rightarrow & 2 \cos 4 x+\left(\frac{1-\cos 4 x}{2}\right)=0 \\
\Rightarrow & 3 \cos 4 x+1=0
\end{array}$
$\begin{array}{ll}\Rightarrow & \cos 4 x=-\frac{1}{3} \\ \Rightarrow & \sin 4 x= \pm \frac{2 \sqrt{2}}{3} \\ \Rightarrow & 4 x=n \pi+(-1)^n \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right) \\ \Rightarrow & x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left( \pm \frac{2 \sqrt{2}}{3}\right)\end{array}$
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