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 $\left[0, 2\pi\right]$ is

Answer 1

$\frac{5}{4}cos^{2} 2x + 1 - 2 \,sin^{2}x\, cos^{2}x + 1 - 3 \,sin^{2}x\, cos^{2}x = 2$
$⇒ 5\, cos^{2}2x - 5 \,sin^{2}2x = 0$
$⇒ 5 \,cos\, 4x = 0$
$⇒ 4x = 2n\pi \pm \frac{\pi}{2}, n ∈ I$
$⇒ x = \frac{n\pi}{2} \pm \frac{\pi}{2}, n ∈ I$
In the interval $\left[0, 2\pi\right]$
Possible solutions are
$\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8}$
So total number of solutions in $\left[0, 2\pi\right]$ is $8$.