Question

The number of distinct solutions of the equation
$\frac{5}{4}co{s}^{2}2x+co{s}^{4}x+si{n}^{4}x+co{s}^{6}x+si{n}^{6}x=2$
in the interval $\left[0,2\pi \right]$ is

Answer 1

Answer: 8

$\frac{5}{4}co{s}^{2}2x+1-2si{n}^{2}xco{s}^{2}x+1-3si{n}^{2}xco{s}^{2}x=2$
$\Rightarrow 5co{s}^{2}2x-5si{n}^{2}2x=0$
$\Rightarrow 5cos4x=0$
$\Rightarrow 4x=2n\pi \pm \frac{\pi }{2},n\in I$
$\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{2},n\in I$
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 $8$.