(2cos²x-√3cosx)/log(4)(sinx)=0
ОДЗ
log(4)sinx≠0⇒sinx≠1⇒cosx≠0
2cos²x-√3cosx=0
cosx*(2cosx-√3)=0
cosx=0 не удов усл
2cosx-√3=0
cosx=√3/2
x=+-π/6+2πn
-3π≤-π/6+2πn≤-3π/2
-18≤-1+12n≤-9
-17≤12n≤-8
-17/12≤n≤-2/3
n=-1⇒x=-π/6-2π=-13π/6
-3π≤π/6+2πn≤-3π/2
-18≤1+12n≤-9
-19≤12n≤-10
-19/12≤n≤-5/6
n=-1⇒x=π/6-2π=-11π/6