Question

16sin^2x-4^sinx / корень(cosx) - 1 =0
очень сроччно

Answer 1

$\frac{16^{Sin^{2}x }-4^{Sinx}}{\sqrt{Cosx} -1} =0\\\\\left \{ {{16^{Sin^{2}x } -4^{Sinx} =0} \atop {\sqrt{Cosx}}-1\neq0 } \right.\\\\\left \{ {{4^{2Sin^{2}x }=4^{Sinx}} \atop {Cosx\geq0;Cosx\neq1}} \right.\\\\2Sin^{2}x=Sinx\\\\2Sin^{2}x-Sinx=0\\\\Sinx(2Sinx-1)=0\\\\1)Sinx=0-neyd\\\\2)2Sinx-1=0\\\\Sinx=\frac{1}{2}\\\\x=(-1)^{n}arcSin\frac{1}{2}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{6}+\pi n,n\in Z$