Number of values of $$x$$ which lie in $$\left[ {0,2\pi } \right]$$  and satisfy the equation $$\left( {\cos \frac{x}{4} - 2\sin x} \right)\sin x + \left( {1 + \sin \frac{x}{4} - 2\cos x} \right)\cos x = 0$$

Solution :
$$\eqalign{ & \left( {\cos \frac{x}{4} - 2\sin x} \right)\sin x + \left( {1 + \sin \frac{x}{4} - 2\cos x} \right)\cos x = 0 \cr & \Rightarrow \left( {\sin x\cos \frac{x}{4} + \cos x\sin \frac{x}{4}} \right) + \cos x - 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0 \cr & \Rightarrow \sin \left( {x + \frac{x}{4}} \right) + \cos x - 2\left( 1 \right) = 0 \cr & \Rightarrow \sin \frac{{5x}}{4} + \cos x = 2 \cr & \Rightarrow \sin \frac{{5x}}{4} = \cos x = 1 \cr & \Rightarrow \sin \frac{{5x}}{4} = 1 \cr & \Rightarrow \frac{{5x}}{4} = 2n\pi + \frac{\pi }{2} \cr & \Rightarrow x = \frac{{8n\pi }}{5} + \frac{{2\pi }}{5}\,\,\& \,\,\cos x = 1 \cr & \Rightarrow x = 2m\pi \cr & {\text{Thus we have, }}\frac{{8n\pi }}{5} + \frac{{2\pi }}{5} = 2m\pi \cr & \Rightarrow m = \frac{{4n + 1}}{5} \cr} $$
$$\therefore n \in I,$$   so $$m$$ must be of the form $$m = 5k + 1$$
Hence, the solution of the equation is
$$x = 2\left( {5k + 1} \right)\pi ,k \in I$$