Question

Помогите решить уравнение

Answer 1

Условие:

$\displaystyle (4cos^2(x)-8cos(x)+3)*\sqrt{8sin(x)} = 0$

ОДЗ:

$\displaystyle sin(x)\geq 0 ~~~~=>\left \{ {{x\geq 0+2\pi n, n\in Z} \atop {x\leq \pi +2\pi n, n\in Z}} \right.$

$\displaystyle (4cos^2(x)-8cos(x)+3)*\sqrt{8sin(x)} = 0 ~~~~=>\left[\begin{array}{c}4cos^2(x)-8cos(x)+3=0\\\sqrt{8sin(x)} = 0\end{array}\right \\\\\\4cos^2(x)-8cos(x)+3=0 ~~~~\bigg| cos(x) = t\bigg| \\\\ 4t^2-8t+3=0\\\\D=64-4*(4*3) = 16\\\\\sqrt{D} = 4\\\\t=\frac{8\pm4}{8} \\\\ t_1 = 1.5\\t_2=0.5$

$\displaystyle \left[\begin{array}{c}cos(x)=1.5\\cos(x)=0.5\end{array}\right\ ~~=>\left[\begin{array}{c}x\in\varnothing\\\\x=\frac{\pi}{3}+2\pi k,k\in Z\\\\x=\frac{5\pi}{3}+2\pi k,k\in Z\end{array}\right$

$sin(x) = 0 => x=\pi n ,n\in Z$

$\displaystyle (x=\frac{5\pi}{3}+2\pi k,k\in Z) \notin ODZ\\\\\text{OTBET} : x= \pi n , n\in Z ~~ or ~~ x = \frac{\pi}{3}+2\pi k, k\in Z$