Question

The solution of trignometric equation
${\cos }^{4}x+{\sin }^{4}x=2\cos (2x+\pi )\cos (2x+\pi )$ is

• A. $x=\frac{n\pi }{2}\pm {\sin }^{-1}\left(\frac{1}{5}\right)$
• B. $x=\frac{n\pi }{4}+\frac{{(-1)}^n}{4}{\sin }^{-1}\left(\frac{1}{5}\right)$
• C. $x=\frac{n\pi }{4}+{\cos }^{-1}\left(\frac{1}{5}\right)$
• D. $x=\frac{n\pi }{4}+\frac{{(-1)}^n}{4}{\cos }^{-1}\left(\frac{1}{5}\right)$
• E. None of the above

Answer 1

The correct option is E None of the above

Given ${\cos }^{4}x+{\sin }^{4}x=2\cos (2x+\pi )\cos (2x+\pi )$
$\Rightarrow {({\cos }^{2}x+{\sin }^{2}x)}^2-2{\cos }^{2}x{\sin }^{2}x=\cos 4x+\cos 2\pi$
$\Rightarrow 1-\frac{1}{2}{\sin }^{2}2x=\cos 4x+1$
$\Rightarrow -\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)=\cos 4x$
$\Rightarrow \cos 4x=-\frac{1}{3}$
$\Rightarrow \sin 4x=\frac{2\sqrt{2}}{3}$
again $\cos 4x=-\frac{1}{3}$
$\Rightarrow 1-2{\sin }^{2}2x=-\frac{1}{3}$
$\sin 2x=\sqrt{\frac{2}{3}}$
Hence, none of the given option is correct.