Question

\sqrt(2cos8x)+3<=3-\sqrt(5sin4x)

Answer 1

$\sqrt{2cos8x}+3\leq3-\sqrt{5sin4x}\\\sqrt{2cos8x}+\sqrt{5sin4x}\leq0\\\boxed{ \left \{ {{\sqrt a\geq0} \atop {\sqrt{2cos8x}+\sqrt{5sin4x}\leq0}} \right.\to\sqrt{2cos8x}+\sqrt{5sin4x}=0 }\\\sqrt{2cos8x}+\sqrt{5sin4x}=0|^2\\2cos8x+5sin4x=0\\2(1-2sin^24x)+5sin4x=0\\2-4sin^24x+5sin4x=0\\4sin^24x-5sin4x-2=0\\\boxed{sin4x=a,|a|\leq1}\\4a^2-5a-2=0\\D=5^2-4*4*(-2)=25+32=57\\a_1= \frac{5-\sqrt{57}}{2*2}\\a_2= \frac{5+\sqrt{57}}{2*2}\to ne\,yd. ysl.(|a_2| \geq1)\\sin4x= \frac{5-\sqrt{57}}{4}$
$x=(-1)^n\,arcsin( \frac{5-\sqrt{57}}{4})+\pi n,n\in z$