The set of values of x satisfying the equatio | vedclass

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(A) Let $	heta = x - \pi/4$. Then $x = 	heta + \pi/4$ and $2x = 2	heta + \pi/2$. Thus,$\cos 2x = \cos(2	heta + \pi/2) = -\sin 2	heta$. The equati

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an empty set

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(A) Let $	heta = x - \pi/4$. Then $x = 	heta + \pi/4$ and $2x = 2	heta + \pi/2$. Thus,$\cos 2x = \cos(2	heta + \pi/2) = -\sin 2	heta$. The equation becomes $2^{	an 	heta} - 2(1/4)^{\frac{\sin^2 	heta}{-\sin 2	heta}} + 1 = 0$. Since $\sin 2	heta = 2 \sin 	heta \cos 	heta$,the exponent is $\frac{\sin^2 	heta}{-2 \sin 	heta \cos 	heta} = -\frac{1}{2} 	an 	heta$. So,$2^{	an 	heta} - 2(2^{-2})^{-\frac{1}{2} 	an 	heta} + 1 = 0$. $2^{	an 	heta} - 2(2^{	an 	heta}) + 1 = 0$. $2^{	an 	heta} - 2 \cdot 2^{	an 	heta} + 1 = 0$ $\Rightarrow 1 - 2^{	an 	heta} = 0$ $\Rightarrow 2^{	an 	heta} = 1$. This implies $	an 	heta = 0$,so $	heta = n\pi$. Thus,$x - \pi/4 = n\pi \Rightarrow x = n\pi + \pi/4$. Since the original equation involves $	an(x - \pi/4)$ and $\cos 2x$ in the denominator,we must check the domain. For $\cos 2x 
eq 0$,$2x 
eq n\pi + \pi/2 \Rightarrow x 
eq n\pi/2 + \pi/4$. For $x = n\pi + \pi/4$,$2x = 2n\pi + \pi/2$,so $\cos 2x = 0$. Since the denominator is zero,the equation is undefined for all $x = n\pi + \pi/4$. Therefore,the set of values is an empty set.

The set of values of $x$ satisfying the equation $2^{\tan(x - \pi/4)} - 2(0.25)^{\frac{\sin^2(x - \pi/4)}{\cos 2x}} + 1 = 0$ is:

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