Question

The number of distinct solutions of the equation
$\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2$
in the interval $[0,2\pi ]$ is

Answer 1

$\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2\Rightarrow \frac{5}{4}{\cos }^{2}2x+({\cos }^{2}x+{\sin }^{2}x{)}^2-2{\cos }^{2}x{\sin }^{2}x+({\cos }^{2}x+{\sin }^{2}x{)}^3-3{\cos }^{2}x{\sin }^{2}x({\cos }^{2}x+{\sin }^{2}x)=2\Rightarrow \frac{5}{4}{\cos }^{2}2x+1-\frac{1}{2}{\sin }^{2}2x+1-\frac{3}{4}{\sin }^{2}2x=2\Rightarrow \frac{5}{4}{\cos }^{2}2x-\frac{5}{4}{\sin }^{2}2x=0\Rightarrow \cos 4x=0=\cos \left(\frac{\pi }{2}\right)\Rightarrow 4x=2n\pi \pm \frac{\pi }{2}\Rightarrow x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}\therefore \text{Number of solutions}=8$