If π/4 less than x less than π/2, then √2 + √2 + 2 cos 4x = .............................. .

If π/4 < x < π/2, Then Find √(2 + √(2 + 2cos4x))

Question:

\[ \frac{\pi}{4}Solution

Using the identity

\[ 1+\cos\theta=2\cos^2\frac{\theta}{2} \]

we get

\[ 2+2\cos4x = 4\cos^2 2x \] \[ \sqrt{2+2\cos4x} = 2|\cos2x| \]

Since

\[ \frac{\pi}{4}Therefore, \(\cos2x<0\), hence

\[ |\cos2x|=-\cos2x \] \[ \sqrt{2+2\cos4x} = -2\cos2x \]

Substituting,

\[ \sqrt{\,2+\sqrt{\,2+2\cos4x\,}} = \sqrt{\,2-2\cos2x\,} \]

Using

\[ 1-\cos2x=2\sin^2x \] \[ 2-2\cos2x = 4\sin^2x \] \[ \sqrt{\,2-2\cos2x\,} = 2|\sin x| \]

Since \(x\) lies in the first quadrant, \(\sin x>0\).

\[ 2|\sin x| = 2\sin x \]

\[ \boxed{2\sin x} \]

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