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

$\frac{5}{4}{\cos }^{2}2x+({\cos }^{2}x{)}^2+({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^3+({\sin }^{2}x{)}^3=2$
$\frac{5}{4}{\cos }^{2}2x+({\cos }^{2}x+{\sin }^{2}x{)}^2-2{\sin }^{2}x\cdot {\cos }^{2}x+({\cos }^{2}x+{\sin }^{2}x\left)\right({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x\cdot {\cos }^{2}x)=2$
$\Rightarrow \frac{5}{4}{\cos }^{2}2x+1-2{\sin }^{2}x.{\cos }^{2}x+({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x\cdot {\cos }^{2}x)=2$
$\frac{5}{4}{\cos }^{2}2x+1-2{\sin }^{2}x.{\cos }^{2}x+(1-3{\sin }^{2}x\cdot {\cos }^{2}x)=2$
$\Rightarrow \frac{5}{4}{\cos }^{2}2x-5{\sin }^{2}x.{\cos }^{2}x=0$
$\frac{5}{4}-\frac{5}{4}{\sin }^{2}2x-5{\sin }^{2}x\cdot {\cos }^{2}x=0$
$\Rightarrow \frac{5}{4}-\frac{5}{4}{\sin }^{2}2x-\frac{5}{4}{\sin }^{2}2x=0$
${\sin }^{2}2x=\frac{1}{2}\Rightarrow \sin 2x=\pm \frac{1}{\sqrt{2}}$
$\therefore x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$
Hence number of solutions in the given interval is $8$.