If cos left(frac{pi}{4}-xright) cos 2 x+sin x | vedclass

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(D) Given equation is: $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \

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$\sqrt{2}$

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(D) Given equation is: $\cos \left(\frac{\pi}{4}-xight) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+xight) \cos 2 x$ Rearranging the terms: $\cos 2 x \left[ \cos \left(\frac{\pi}{4}-xight) - \cos \left(\frac{\pi}{4}+xight) ight] = \sin 2 x \sec x (\cos x - \sin x)$ Using the formula $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$: $\cos 2 x \left[ 2 \sin \frac{\pi}{4} \sin x ight] = \sin 2 x \sec x (\cos x - \sin x)$ Since $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin 2 x = 2 \sin x \cos x$: $\cos 2 x \left( \frac{2}{\sqrt{2}} \sin x ight) = (2 \sin x \cos x) \sec x (\cos x - \sin x)$ $\sqrt{2} \cos 2 x \sin x = 2 \sin x (\cos x - \sin x)$ Assuming $\sin x 
eq 0$,we divide by $\sin x$: $\sqrt{2} (\cos^2 x - \sin^2 x) = 2 (\cos x - \sin x)$ $\sqrt{2} (\cos x - \sin x)(\cos x + \sin x) = 2 (\cos x - \sin x)$ If $\cos x - \sin x 
eq 0$,then $\sqrt{2} (\cos x + \sin x) = 2$,which implies $\cos x + \sin x = \sqrt{2}$. Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = 1$,which is $\cos \left(x - \frac{\pi}{4}ight) = 1$. Thus,$x - \frac{\pi}{4} = 0$,so $x = \frac{\pi}{4}$. Then $\sec x = \sec \frac{\pi}{4} = \sqrt{2}$.

If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$,then a possible value of $\sec x$ is

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