Official solution
\[2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1\]
\[2\cos x\left( {4\left( {{{\sin }^2}{\pi \over 4} - {{\sin }^2}x} \right) - 1} \right) = 1\]
\[2\cos x\left( {4\left( {{1 \over 2} - {{\sin }^2}x} \right) - 1} \right) = 1\]
\[2\cos x(2 - 4{\sin ^2}x - 1) = 1\]
\[2\cos x(1 - 4{\sin ^2}x) = 1\]
\[2\cos x(4{\cos ^2}x - 3) = 1\]
\[4{\cos ^3}x - 3\cos x = {1 \over 2}\]
\[\cos 3x = {1 \over 2}\]
\[x \in [0,\pi ]\] \[\therefore\] \[3x \in [0,3\pi ]\]