Question

(4cos^2x-4cosx-3)*log14(-sinx)=0

Answer 1

$left[ begin{matrix} begin{cases} 4cos^2x-4cos x-3=0 \ -sin x>0 end{cases} \ log_{14}(-sin x)=0 end{matrix}right. <=> left[ begin{matrix} begin{cases} 4cos^2x-4cos x-3=0 \ sin x<0 end{cases} \ begin{cases} sin x=-1 \ sin x<0 end{cases} end{matrix}right. <=>$
$left[ begin{matrix} begin{cases} left[ begin{matrix} cos x=-0,5\ cos x=1,5 end{matrix}right \ sin x<0 end{cases} \ sin x=-1 end{matrix}right. => begin{cases} sin x<0 \ left[ begin{matrix} cos x=-0,5 \ sin x=-1 end{matrix}right end{cases}. <=>$
$begin{cases} sin x<0 \ left[ begin{matrix} x=pm frac{2pi}{3}+2pi k \ x=-frac{pi}{2}+2pi n end{matrix}right end{cases} =>left[ begin{matrix} x=-frac{2pi}{3}+2pi k \ x=-frac{pi}{2}+2pi n end{matrix}right\ Ombem: -frac{2pi}{3}+2pi k; -frac{pi}{2}+2pi n; k,n in Z$