Question

Number of roots the equation $2^{\tan (x-\frac{\pi }{4})}-2(0.25{)}^{\frac{{\sin }^2(x-\frac{\pi }{4})}{\cos 2x}}+1=0$ is

Answer 1

$2^{\tan (x-\frac{\pi }{4})}-2(0.25{)}^{\frac{{\sin }^2(x-\frac{\pi }{4})}{\cos 2x}}+1=0$
$2^{\tan (x-\frac{\pi }{4})}-2(0.5{)}^{\frac{2{\sin }^2(x-\frac{\pi }{4})}{\cos 2x}}+1=0$
$2^{\tan (x-\frac{\pi }{4})}-2(0.5{)}^{\frac{1-\cos (2x-\frac{\pi }{2})}{\cos 2x}}+1=0$
$2^{\tan (x-\frac{\pi }{4})}-2(0.5{)}^{\frac{1-\sin 2x}{\cos 2x}}+1=0$
$2^{\tan (x-\frac{\pi }{4})}-2(0.5{)}^{\frac{\cos x-\sin x}{\cos x+\sin x}}+1=0$
$2^{\tan (x-\frac{\pi }{4})}-2(0.5{)}^{-\tan (x-\frac{\pi }{4})}+1=0$
$2^{\tan (x-\frac{\pi }{4})}=1$
$x=n\pi +\frac{\pi }{4}$
But we must also have constraint that, $\cos 2x\ne 0$
Thus there is no solution for the given equation.