Решите, пожалуйста, 2 примера: 3cos^2 x+4sinx=0 log x > 1 1/2

13cos^2 (x)+4sin (x)=03*(1-sin^2 (x)+4sin (x)=03sin^2 (x) -4sin (x) -3=0sin (x)=t3t^2-4t-3=0D=b^2-4ac=16+48=52t1,2=(-b±sqrt (D) /2*at1,2=(4±sqrt (52) /6t1=(4+sqrt (52) /6=(2+sqrt (13) /3t2=(2sqrt (13) /31) sin (x)=(2+sqrt (13) /32) sin (x)=(2-sqrt (13) /3 x=(-1) ^n*arcsin (2+sqrt (13) /3)+pi*n x=(-1) ^n*arcsin (2-sqrt (13) /3)+pi*n 2log (1/2; x) >1log (1/2; x) >log (1/2; 1/2) x<1/2 и x>0