Question

The solution(s) of the equation $9{\cos }^{12}x+{\cos }^{2}2x+1=6{\cos }^{6}x\cos 2x+6{\cos }^{6}x-2\cos 2x$ is/are $(n\in I)$.

• A. $n=n\pi +\frac{\pi }{2}$
• B. $n=n\pi +{\cos }^{-1}\left(\sqrt[4]{4\frac{2}{3}}\right)$
• C. $n=n\pi -{\cos }^{-1}\left(\sqrt[4]{4\frac{2}{3}}\right)$
• D. $n=n\pi +{\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $n=n\pi +\frac{\pi }{2}$
B $n=n\pi +{\cos }^{-1}\left(\sqrt[4]{4\frac{2}{3}}\right)$
C $n=n\pi -{\cos }^{-1}\left(\sqrt[4]{4\frac{2}{3}}\right)$

$(3{\cos }^{6}x-1-\cos 2x{)}^2=0$
$\Rightarrow {\cos }^{2}x(3{\cos }^{4}x-2)=0$
$\Rightarrow \cos x=0\text{or}{\cos }^{4}x=\frac{2}{3}$
$\Rightarrow n=n\pi +\frac{\pi }{2}\text{or}n=n\pi \pm {\cos }^{-1}\left(\sqrt[4]{4\frac{2}{3}}\right)$