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

1

3cos^2(x)+4sin(x)=0

3*(1-sin^2(x)+4sin(x)=0

3sin^2(x)-4sin(x)-3=0

sin(x)=t

3t^2-4t-3=0

D=b^2-4ac=16+48=52

t1,2=(-b±sqrt(D))/2*a

t1,2=(4±sqrt(52)/6

t1=(4+sqrt(52))/6=(2+sqrt(13))/3

t2=(2sqrt(13))/3

1)sin(x) =(2+sqrt(13))/3

2)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

2

log(1/2;x)>1

log(1/2;x)>log(1/2;1/2)

x<1/2  и x>0