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 $[0,2\pi ]$ is ___

Answer 1

$\frac{5}{4}co{s}^{2}2x+co{s}^{4}x+si{n}^{4}x+co{s}^{6}x+si{n}^{6}x=2\Rightarrow \frac{5}{4}co{s}^{2}x+1-\frac{1}{2}si{n}^{2}2x+1-\frac{3}{4}si{n}^{2}2x=2\Rightarrow \frac{5}{4}(co{s}^{2}2x-si{n}^{2}2x)=0\Rightarrow cos4x=0$
$\Rightarrow 4x=(2n+1)\frac{\pi }{2}$ or $x=(2n+1)\frac{\pi }{8}$
For $xϵ[0,2\pi ]$, n can take values 0 to 7
$\therefore$ 8 solutions.